Solucionario Fundamentos de Física 9na edición Capitulo 3

3
Vectors and Two-Dimensional Motion
CLICKER QUESTIONS
Question A3.01
Description: Understanding acceleration: identifying nonzero acceleration.
Question
Consider the following situations:
• A car slowing down at a stop sign
• A ball being swung in a circle at constant speed
• A vibrating string
• The Moon orbiting the Earth
• A skydiver falling at terminal speed
• An astronaut in an orbiting space station
• A ball rolling down a hill
• A person driving down a straight section of highway at constant speed with her foot on the accelerator
• A molecule in the fl oor of this room
In how many of the situations is the object accelerating?
1. One of the situations
2. Two of the situations
3. Three of the situations
4. Four of the situations
5. Five of the situations
6. Six of the situations
7. Seven of the situations
8. Eight of the situations
9. Nine of the situations
10. None of the situations
Commentary
Purpose: To introduce the concept of acceleration, explore your preexisting defi nitions of the term, and
contrast “everyday” usage of the word with the way physicists use it.
Discussion: In physics, “acceleration” means the rate of change of velocity with respect to time. The more
the velocity changes during a certain interval (say, 1 second), the larger the acceleration is.
75

76 Chapter 3
There are three ways in which the velocity can change: (1) the speed increases; (2) the speed decreases; or
(3) the direction of motion changes. If any one of these occurs, then the object is said to be “accelerating.”
In other words, unlike everyday usage, “accelerating” does not necessarily mean “speeding up.”
When a car is slowing down, we say it is accelerating. We generally avoid the phrases “deceleration” and
“negative acceleration” as unnecessary and unhelpful. We avoid “negative acceleration” because an object
can have a negative value for acceleration and still be speeding up—for example, if it is moving in the
negative direction.
When a ball is moving in a circle, its direction of motion is constantly changing, so it is accelerating even if
its speed is constant.
A vibrating string (except for its endpoints) is accelerating at most instants in time, though it happens so
quickly that we cannot perceive the changes in velocity. The only time it is not accelerating is when its
speed is maximum. When the string stops and reverses direction at the extremes of its motion, its accelera-tion
is largest.
The Moon orbiting the Earth is moving in a circle, so it is accelerating.
A skydiver moving at terminal speed has a constant velocity and thus zero acceleration.
An astronaut in an orbiting space station may not be moving relative to the space station, but the station is
not a proper frame of reference since it is accelerating as it orbits the Earth. (In this course, a proper frame
of reference has no acceleration.) The Earth is a better frame of reference, and the astronaut is moving in a
circle around it, so she is accelerating.
A ball rolling down a hill generally speeds up.
A person (or car, or any object) moving at constant speed along a straight line has constant velocity, and
thus zero acceleration. This is true even though we must press the “accelerator” pedal to maintain constant
velocity.
A molecule in a solid that is not at absolute zero temperature is vibrating, so like the vibrating string it is
accelerating most of the time. The solid as a whole may still be at rest.
Key Points:
• An object’s acceleration is the rate at which its velocity is changing.
• If an object’s speed or direction of motion changes, we say it is “accelerating.”
• We avoid using the phrases “deceleration” and “negative acceleration” to mean “slowing down.”
For Instructors Only
A nice feature of this style of question (“how many of the following . . .”) is that even if a student answers
the “correct” number of objects (7), there are 36 different combinations to arrive at this answer, so there is
really no need to focus any attention at all on the correctness of students’ multiple-choice answers. Rather,
class discussion should proceed immediately to considering each situation independently and applying the
defi nition of acceleration.
We encourage students to wrestle with their own defi nitions at this stage. This question may be used prior
to any formal presentation of acceleration, since everyone already “knows” the term; the disagreements that
typically arise prepare students to appreciate and better understand the formal defi nition.

Vectors and Two-Dimensional Motion 77
It is not too early to mention forces, at least informally. Many students already have some ideas about
force that can help them sort out what you are trying to explain regarding acceleration. For example, in
the cases that have nonzero acceleration, identify the force(s) causing it; in cases with zero acceleration,
discuss how any forces present balance out.
Question A3.02
Description: Develop understanding of vector acceleration in curvilinear motion.
Question
A pendulum is released from rest at position A and swings toward the vertical under the infl uence of
gravity as depicted below.
A
B
8
7
1
6 2
5 3
4
C
When at position B, which direction most nearly corresponds to the direction of the acceleration?
Enter (9) if the direction cannot be determined.
Commentary
Purpose: Develop your understanding of the vector nature of acceleration in curvilinear motion.
Description: Acceleration is the rate of change of velocity. If an object’s speed is changing, that
means the length of its velocity vector is changing, so the acceleration must have a nonzero component
parallel to the velocity vector. If an object’s direction of motion is changing, that means the direction of
the velocity vector is changing, so the acceleration must have a nonzero component perpendicular to the
velocity vector.
At point B, the pendulum bob is both speeding up and changing direction. At that moment its velocity
points in a direction tangential to the circular path it follows: direction (3). Since it is speeding up, the
acceleration must have a component that points in direction (3). Since it is traveling in a curved path, it
must have an acceleration component that points towards the center of the curve (the so-called “centrip-etal
acceleration”), direction (1). Thus, the acceleration vector must point somewhere between directions
(1) and (3), so (2) is probably the best choice.
(To fi nd the exact direction, we would have to calculate the relative sizes of these two components of the
acceleration.)

78 Chapter 3
Key Points:
• Acceleration is a vector that describes the rate of change of the velocity vector’s magnitude and its
direction.
• It is often useful to divide the acceleration into components that are parallel (tangential) and
perpendicular to the object’s direction of motion.
• A nonzero tangential component of acceleration indicates that the object is changing speed.
• A nonzero perpendicular component of acceleration indicates that the object is changing direction.
For Instructors Only
Students often neglect one component or the other in this question. Common answers include (1) (fi xated
on the centripetal acceleration or the string), (3) (fi xated on tangential acceleration and the increase in
speed), and (4) (fi xated on gravity).
Students choosing answer (5) might be revealing a misunderstanding of the fi ctitious “centrifugal force.”
Answer (9), impossible to determine, is defensible if chosen because students don’t know enough to
determine the relative sizes of the two acceleration components and therefore can’t choose among
directions (1), (2), and (3).
QUICK QUIZZES
1. (c). The largest possible magnitude of the resultant occurs when the two vectors are in the same

and B
direction. In this case, the magnitude of the resultant is the sum of the magnitudes of A

:
R = A + B = 20 units. The smallest possible magnitude of the resultant occurs when the two vec-tors

and B
are in opposite directions, and the magnitude is the difference of the magnitudes of A

:
R = |A − B| = 4 units.
2. Vector x-component y-component
− +
A
+ −
B

+ B
A
− −
3. (b). If velocity is constant, the acceleration (rate of change in velocity) is zero. An object may
have constant speed (magnitude of velocity) but still be accelerating due to a change in direction
of the velocity. If an object is following a curved path, it is accelerating because the velocity is
changing in direction.
4. (a). Any change in the magnitude and/or direction of the velocity is an acceleration. The gas pedal
and the brake produce accelerations by altering the magnitude of the velocity. The steering wheel
produces accelerations by altering the direction of the velocity.
5. (c). A projectile has constant horizontal velocity. Thus, if the runner throws the ball straight up
into the air, the ball maintains the horizontal velocity it had before it was thrown (that is, the
velocity of the runner). In the runner’s frame of reference, the ball appears to go straight upward
and come straight downward. To a stationary observer, the ball follows a parabolic trajectory,
moving with the same horizontal velocity as the runner and staying above the runner’s hand.

6. (b). The velocity is always tangential to the path while the acceleration is always directed
vertically downward. Thus, the velocity and acceleration are perpendicular only where the
path is horizontal. This only occurs at the peak of the path.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
1. At maximum height (Δy = hmax), the vertical velocity of the stone will be zero. Thus,
= 2 + 2 (Δ ) gives
v v y y y 2 a y
0
h
v 2
v 2
v −( ) y y
2 2 2
a g
y
max
sin s
=
−
=
−
(− ) =
0
0
2
0
2
θ 45 m s in .
.
.
2 55 0
2 980
69 3
( °)
(− ) =
m s
m 2
and we see that choice (c) is the correct answer.
2. The skier has zero initial velocity in the vertical direction (v ) 0 0 y = and undergoes a vertical
displacement of Δy = −3.20 m. The constant acceleration in the vertical direction is a = − g , so
y we use Δy = v t + 1
at 2 to fi nd the time of fl ight as
0
y 2
y −3 20 = 0 + (− )
1
2
. m 9.80 m s2 t 2 or t =
(− )
−
=
2 320
9 80
0 808
.
.
.
m
m s
s 2
During this time, the object moves with constant horizontal velocity v v x x = = 0 22 0. m s. The
horizontal distance traveled during the fl ight is
Δx tx = v = (22.0 m s)(0.808 s) = 17.8 m
which is choice (d).
3. Choose coordinate system with north as the positive y-direction and east as the positive
x-direction. The velocity of the cruise ship relative to Earth is v
CE = 4.50 m s due north, with
components of ( ) ( ) . v v CE x CE y = 0 and = 4 50 m s. The velocity of the patrol boat relative to
Earth is v
PE = 5.20 m s at 45.0° north of west, with components of
v v PE PE ( ) = − ° = −( )( )= − x cos45.0 5.20 m s 0.707 3.68 m s
and
v v PE PE ( ) = + °= ( )( ) = y sin 45.0 5.20 m s 0.707 3.68 m s
Thus, the velocity of the cruise ship relative to the patrol boat is v v v CP CE PE = − , which has
components of
v v v CP CE PE ( ) = ( ) − ( ) = − (− ) = + x x x 0 3.68 m s 3.68 m s
and
v v v CP CE PE ( ) = ( ) − ( ) = − = + y y y 4.50 m s 3.68 m s 0.823m s
Choice (a) is then the correct answer.

80 Chapter 3
4. For vectors in the x-y plane, their components their components have the signs indicated in the
following table:
Quadrant of Vector
First Second Third Fourth
x-component Positive Negative Negative Positive
y-component Positive Positive Negative Negative
Thus, a vector having components of opposite sign must lie in either the second or fourth
quadrants and choice (e) is the correct answer.
5. The path followed (and distance traveled) by the athlete is
shown in the sketch, along with the vectors for the initial
position, fi nal position, and change in position.
The average speed for the elapsed time interval Δt is
vav = d
Δt
and the magnitude of the average velocity for this time
interval is

v
r
av =
Δ
Δt
The sketch clearly shows that d Δr
in this case, meaning that vav av v
and that (a) is the
correct choice.
6. Consider any two very closely spaced points on a circular path and draw vectors of the same
length (to represent a constant velocity magnitude or speed) tangent to the path at each of these
points as shown in the leftmost diagram below. Now carefully move the velocity vector v
f at
the second point down so its tail is at the fi rst point as shown in the rightmost diagram. Then,
draw the vector difference Δv = v − v f i and observe that if the start of this vector were located on
the circular path midway between the two points, its direction would be inward toward the center
of the circle.
Thus, for an object following the circular path at constant speed, its instantaneous acceleration,
a= lim
( Δ vΔ
t
) Δ
t
→
0
, at the point midway between your initial and end points is directed toward the
center of the circle, and the only correct choice for this question is (d).

7. Whether on Earth or the Moon, a golf ball is in free fall with a constant downward acceleration of
magnitude determined by local gravity from the time it leaves the tee until it strikes the ground or
other object. Thus, the vertical component of velocity is constantly changing while the horizontal
component of velocity is constant. Note that the speed (or magnitude of the velocity)
v v v = = + v
2 2
x y
will change in time since vy changes in time. Thus, the only correct choices for this question are
(b) and (d).
8. At maximum altitude, the projectile’s vertical component of velocity is zero. The time for the
projectile to reach its maximum height is found from v v y y y = + a t 0 as
t
−
v v y Δ y h 0
y
v v 0 0 0 0 0 θ θ
a g g
y
max
=
sin sin max =
−
−
= =
Since the acceleration of gravity on the Moon is one-sixth that of Earth, we see that
(with v0 0 and θ kept constant)
t
θ θ θ0 6
= = = v v v 0 0 0 0 0
sin sin sin
max
Moon
g g Moon Earth
6
6
g
t
Earth
Eart
⎛
⎝ ⎜
⎞
⎠ ⎟
= max
and the correct answer for this question is (e).
9. The boat moves with a constant horizontal velocity (or its velocity relative to Earth has
components of ( ) ( ) v v BE BE constant, and x y = v = = 0 0), where the y-axis is vertical and the x-axis
is parallel to the keel of the boat. Once the wrench is released, it is a projectile whose velocity
relative to Earth has components of
v
WE ( ) = + = + = x x x v a t v v 0 0 0 0 and v
WE ( ) = + = − =− y y y v a t gt gt 0 0
The velocity of the wrench relative to the boat ( ) v v v WB WE BE = − has components of
v v v WB WE BE ( ) = ( ) − ( ) = − =
x x x v v 0 0 0 and v v v WB WE BE ( ) = ( ) − ( ) = − − = − y y y gt 0 gt
Thus, the wrench has zero horizontal velocity relative to the boat and will land on the deck at a
point directly below where it was released (i.e., at the base of the mast). The correct choice is (b).
10. While in the air, the baseball is a projectile whose velocity always has a constant horizontal
component (v v ) x x = 0 and a vertical component that changes at a constant rate (Δv Δ ) y y t = a = −g .
At the highest point on the path, the vertical velocity of the ball is momentarily zero. Thus,
at this point, the resultant velocity of the ball is horizontal and its acceleration continues to be
directed downward (a , a g) x y = 0 = − . The only correct choice given for this question is (c).

82 Chapter 3
11. Note that for each ball, v0 0 y = , Thus, the vertical velocity of each ball when it reaches the
= 2 + 2 (Δ ) as
ground (Δy = −h) is given by v v y y y 2 a y
0
vy = − 0 + 2(−g)(−h) = − 2gh
and the time required for each ball to reach the ground is given by v v y y y = + a t 0 as
t
v v0 2 0 2
a
gh
g
h
g
y y
y
=
−
=
− −
−
=
The speeds (i.e., magnitudes of total velocities) of the balls at ground level are
Red Ball: v = v 2 + v 2 = v 2 + (− 2 gh ) 2
= v 2 + 2
gh
R x y 0
x 0
Blue Ball: v = v 2 + v 2 = v 2 + (− ) 2
2 gh = 0 + 2 gh = 2
gh
B x y 0
x Therefore, we see that the two balls reach the ground at the same time but with different speeds
(v v ) R B , so only choice (b) is correct.
12. When the apple fi rst comes off the tree, it is moving forward with the same horizontal velocity as
the truck. Since, while in free fall, the apple has zero horizontal acceleration, it will maintain this
constant horizontal velocity as it falls. Also, while in free fall, the apple has constant downward
acceleration (a g) y = − , so its downward speed increases uniformly in time.
(i) As the truck moves left to right past an observer stationary on the ground, this observer will
see both the constant velocity horizontal motion and the uniformly accelerated downward
motion of the apple. The curve that best describes the path of the apple as seen by this
observer is (a).
(ii) An observer on the truck moves with the same horizontal motion as does the apple. This
observer does not detect any horizontal motion of the apple relative to him. However,
this observer does detect the uniformly accelerated vertical motion of the apple. The curve
best describing the path of the apple as seen by the observer on the truck is (b).
13. Of the choices listed, the quantities that have magnitude or size, but no direction, associated with
them (i.e., scalar quantities) are (b) temperature, (c) volume, and (e) height. The other quantities,
(a) velocity of a sports car and (d) displacement of a tennis player who moves from the court’s
backline to the net, have both magnitude and direction associated with them, and are both vector
quantities.
ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS
2. The components of a vector will be equal in magnitude if the vector lies at a 45° angle with the two
axes along which the components lie.
4. The minimum sum for two vectors occurs when the two vectors are opposite in direction. If they are
unequal, their sum cannot be zero.
6. The balls will be closest at the instant the second ball is projected. The fi rst ball will always be going
faster than the second ball. There will be a one second time interval between their collisions with
the ground. The two move with the same acceleration in the vertical direction. Thus, changing their
horizontal velocity can never make them hit at the same time.

8. The equations of projectile motion are only valid for objects moving freely under the infl uence of
gravity. The only acceleration such an object has is the free-fall acceleration, g, directed vertically
downward. Of the objects listed, only a and d meet this requirement.
10. The passenger sees the ball go into the air and come back in the same way he would if he were at rest
on Earth. An observer by the tracks would see the ball follow the path of a projectile. If the train were
accelerating, the ball would fall behind the position it would reach in the absence of the acceleration.
PROBLEM SOLUTIONS
3.1 We are given that R A B

= + . When two vectors are added graphically, the
second vector is positioned with its tail at the tip of the fi rst vector. The resultant
then runs from
the tail of the fi rst vector to the tip of the second vector. In this
case, vector A
will be positioned with its tail at the origin and its tip at the
point (0, 29). The resultant is then drawn, starting at the origin (tail of fi rst
vector) and going 14
units in the negative y-direction to
the point (0, −14).
The second vector, B
, must then start from the tip of A
at point (0, 29) and

at point (0, −14) as shown in the sketch at the right.
end on the tip of R
From this, it is seen that

is 43 units in the negative y-direction
B
3.2 (a) Using graphical methods, place the

at the head of
tail of vector B

. The new vector A B
vector A

+ has
a magnitude of 6.1 units at 113°
from the positive x-axis.

− is found
(b) The vector difference A B

by placing the negative of vector B
(a vector of the same magnitude as
B

. The resultant vector A B

, but opposite direction) at the head of vector A

− has
magnitude 15 units at 23° from the positive x-axis.

at the
3.3 (a) In your vector diagram, place the tail of vector B

. The vector sum, A B
tip of vector A

+ , is then found as
shown in the vector diagram and should be
A B

+ = 5.0 units at − 53°

(same
(b) To fi nd the vector difference, form the vector −B

, opposite direction) and add it to vector
magnitude as B

as shown in the diagram. You should fi nd that
A B
A

− = 5.0 units at + 53°

84 Chapter 3
3.4 Sketches of the scale drawings needed for parts (a) through (d) are given below. Following the
sketches is a brief comment on each part with its answer.
(a) (b) (c) (d)
(a) Drawing the vectors to scale and maintaining their respective directions yields a resultant
of 5.2 m at+ 60° .

, but reverse the direction of B
(b) Maintain the direction of A

. The resultant

to produce −B
is 3.0 m at − 30° .

, but reverse the direction of A
(c) Maintain the direction of B

. The resultant
to produce −A
is 3.0 m at + 150° .

, reverse the direction of B
(d) Maintain the direction of A

and double its magnitude, to

. The resultant is 5.2 m at − 60° .
produce −2B

3.5 Your sketch should be drawn to scale, similar to that pictured below. The length of R
and the
angle θ can be measured to fi nd, with use of your scale factor, the magnitude and direction of the
resultant displacement. The result should be
→
approximately 421 ft at 3° below the horizontal
3.6 (a) The distance d from A to C is
d = x + y 2 2
where x = 200 km + (300 km)cos30.0° = 460 km
and y = 0 + (300 km)sin 30.0° = 150 km
∴ d = (460 km)2 + (150 km)2 = 484 km
(b) φ = ⎛⎝ ⎜
⎞⎠ ⎟
= ⎛⎝ ⎜
⎞⎠ ⎟
tan−1 tan−1 150 = 18
.
y
x
km
460 km
1° N of W
(c) Because of the curvature of the Earth, the plane doesn’t travel along straight lines . Thus,
the answer computed above is only approximately correct.

3.7 Using a vector diagram, drawn to scale, like that shown
at the right,
the fi nal displacement of the plane can be
found to be R
plane = 310 km at θ = 57° N of E. The
requested displacement of the base from point B is
−R

plane, which has the same magnitude but the
opposite direction. Thus, the answer is

plane 310 km atθ = 57° S ofW
−R =
3.8 Your vector diagram should look like the
one shown at the right. The initial
displacement A

= 100 m due west and the

= 175 m at 15.0° N of W are
resultant R
both known. In order to reach the end point
of the run following the initial displacement,

the jogger must follow the path shown as B
.

and the angle θ can be
The length of B
measured. The results should be 83 m at 33° N ofW .

= 8.00 m westward and
3.9 The displacement vectors A

B
= 13.0 m north can be drawn to scale as at the right.
The vector C

represents the displacement that the man
in the maze must undergo to return to his starting point.
The scale used to draw the sketch can be used to
fi nd C

to be 15 m at 58° S of E .

are its projections
3.10 The x- and y-components of vector A
on lines parallel to the x- and y-axis, respectively, as shown
in the sketch. The magnitude of these components can be
computed using the sine and cosine functions as shown below:

A= A cos325 ° = + A cos35 ° = ( 35.0 ) cos35 ° =
28.7 units
xand

sin 325 sin 35 35.0 sin 35 20.1 units
Ay= A ° = − A ° = −( ) ° = −
3.11 Using the vector diagram given at the right, we fi nd
R = (6 00 ) + (5 40 ) = 8 07 2 2 . m . m . m
and
θ = ⎛⎝ ⎜
⎞⎠ ⎟
1 tan 1 . . 5 40
6 00
.
.
tan− = − ( ) =
0 900 42 0
m
m
°
Thus, the required displacement is 8.07 m at 42.0° S of E .
plane
→

86 Chapter 3
3.12 (a) The skater’s displacement vector, d
, extends in a straight
line from her starting point A to the end point B. When
she has coasted half way around a circular path as shown
in the sketch at the right, the displacement vector
coincides with the diameter of the circle and has
magnitude
d
= 2r = 2(5.00 m) = 10.0 m
C
r5.00 m
d
B A
(b) The actual distance skated, s, is one half the circumference of the circular path of radius r.
Thus,
s = ( r) = ( ) = 1
2
2π π 5.00 m 15.7 m
(c) When the skater skates all the way around the circular path, her end point, B, coincides
with the start point, A. Thus, the displacement vector has zero length, or
d
= 0
3.13 (a) Her net x (east–west) displacement is − 3.00 + 0 + 6.00 = + 3.00 blocks, while her net y
(north–south) displacement is 0 + 4.00 + 0 = + 4.00 blocks. The magnitude of the resultant
displacement is
R = (Σx) + (Σy) = ( ) + ( ) = 2 2 2 2 3.00 4.00 5.00 blocks
and the angle the resultant makes with the x-axis (eastward direction) is
θ =
⎛
1 1 tan 1 . 4 00
⎝ ⎜
⎞
⎠ ⎟
= ⎛⎝
⎞⎠
.
.
tan− tan− = −
3 00
1
Σ
Σ
y
x
( 33) = 53.1°
The resultant displacement is then 5.00 blocks at 53.1° N of E .
(b) The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks .

= + + + , where A
3.14 (a) The resultant displacement is R A B C D

= 250 m

= 75 0 . m due north, B

= 125 m at 30.0° north of east, and D
due east, C

= 150 m due south. Choosing east as the
positive x-direction and north as the positive y-direction, we fi nd the components of the
resultant to be
R A B C D x x x x x = + + + = 0 + 250 m + (125 m)cos30.0° + 0 = 358 m
and
R A B C D y y y y y = + + + = 75.0 m + 0 + (125 m)sin 30.0° + (−150 m) = −12.5 m
The magnitude and direction of the resultant are then
R R R x y = 2 + 2 = ( )2 + (− )2 = 358 m 12.5 m 358 m
and
θ =
⎛
1 1 12.5
⎝ ⎜
⎞
⎠ ⎟
= − ⎛⎝ ⎜
⎞⎠ ⎟
tan− tan− = −
358
R
R
y
x
m
m
2.00°

= 358 m at 2.00° south of east .
Thus, R
(b) Because of the commutative property of vector addition, the net displacement is the same
regardless of the order in which the individual displacements are executed.

3.15 Ax = −25 0 . Ay = 40 0 .
A = Ax2 + A2 = (− 25.0 )2 + ( 40.0 )2 = 47.2 units
yFrom the triangle, we fi nd that
⎛
φ =
1 1 40 0
. ⎝ ⎜
⎞
⎠ ⎟
= ⎛⎝
⎞⎠
.
.
tan− tan− = °
25 0
58 0
A
A
y
x
, so θ = 180° −φ = 122°

= 47.2 units at 122° counterclockwise from +x-axis .
Thus, A

3.16 Let A

be the vector corresponding to the 10.0-yd run, B
to the 15.0-yd run, and
C
to the
50.0-yd pass. Also, we choose a coordinate system with the + y direction downfi eld, and the
+ x direction toward the sideline to which the player runs.
The components of the vectors are then
Ax = 0
Ay = −10 0 . yds
Bx = 15 0 . yds
By = 0
Cx = 0 Cy = +50 0 . yds
From these, R x x= Σ = 15.0 yds, and R y y= Σ = 40.0 yds, and
R R R x y = 2 + 2 = ( )2 + ( )2 = 15.0 yds 40.0 yds 42.7 yards
3.17 After 3.00 h moving at 41.0 km/h, the hurricane is 123 km at 60.0° N of W from the island. In
the next 1.50 h, it travels 37.5 km due north. The components of these two displacements are as
follows:
Displacement x-component (eastward) y-component (northward)
123 km - 61.5 km +107 km
37.5 km 0 +37.5 km
Resultant - 61.5 km 144 km
Therefore, the eye of the hurricane is now
R = (− 61 5 ) + (144 ) = 2 2 . km km 157 km from the island

88 Chapter 3
3.18 Choose the positive x-direction to be eastward and positive y as northward. Then, the components
of the resultant displacement from Dallas to Chicago are
Rx = Σx = (730 mi)cos 5.00° − (560 mi)sin 21.0° = 527 mi
and
R y y= Σ = (730 mi)sin 5.00° + (560 mi)cos 21.0° = 586 mi
R = R 2 + R 2 = ( 527 mi )2 + ( 586 mi )2 = 788 mi
x y ⎛
θ =
tan−1 = tan−1 (1.11) = 48.1°
⎝ ⎜
⎞
⎠ ⎟
Σ
Σ
y
x
Thus, the displacement from Dallas to Chicago is
R

= 788 mi at 48.1° N of E .
3.19 The components of the displacements a
,
b, and c
are
a a x = ⋅ cos30.0° = +152 km
b b x = ⋅ cos110° = − 51.3 km
c c x = ⋅ cos180° = −190 km
and
a a y = ⋅ sin 30.0° = +87.5 km
b b y = ⋅ sin110° = +141 km
c c y = ⋅ sin180° = 0
Thus,
R a b c x x x x = + + =−89.7 km, and R a b c y y y y = + + = + 228 km
so
R R R x y = 2 + 2 = 245 km, and θ =
⎛
tan−1 = tan−1 (1.11) = 21.4°
⎝ ⎜
⎞
⎠ ⎟
R
R
x
y
City C is 245 km at 21.4°W of N from the starting point.
(north)
(east)

3.20 (a) F1 F1x = 120 N = (120 N)cos60.0° = 60.0 N F N N 1y = (120 )sin 60.0° = 104
F F2 2x = 80.0 N = −(80.0 N)cos 75.0° = − 20.7 N F N N 2y = (80.0 )sin 75.0° = 77.3
F F F R x y = (Σ ) + (Σ ) = ( ) + ( ) = 2 2 2 2 39.3 N 181 N 185 N
and θ = ⎛⎝ ⎜
⎞⎠ ⎟
1 tan 1 . . 181
39 3
tan− = − ( ) = °
.
4 61 77 8
N
N

= 185 N at 77.8° from the x-axis
The resultant force is F
R
(b) To have zero net force on the mule, the resultant above must be cancelled by a force equal
in magnitude and oppositely directed. Thus, the required force is
185 N at 258° from the x-axis
3.21 The single displacement required to sink the putt in one stroke is equal to the resultant of
the three actual putts used by the novice. Taking east as the positive x-direction and north
as the positive y-direction, the components of the three individual putts and their resultant are
A= 0 A= +4.00 m
x y B= (2.00 m)cos 45.0° = +1.41 m
B= (2.00 m)sin 45.0° = +1.41 m
x y Cx = −(1.00 m)sin 30.0° = −0.500 m Cy = −(1.00 m)cos30.0° = −0.866 m
R A B C x x x x = + + = + 0.914 m R A B C y y y y = + + = +4.55 m
The magnitude and direction of the desired resultant is then
R R R x y = 2 + 2 = 4.64 m and θ =
⎛
⎝ ⎜⎞
tan−1 = +78.6°
⎠ ⎟
R
R
y
x
Thus, R = 4.64 m at 78.6° north of east .
45 1 x = ( )⎛
3.22 v0 101
0 477
1
mi h =
⎝ ⎜
⎞
⎠ ⎟
m s
mi h
m s
.
. and Δx = ( )⎛⎝ ⎜
⎞⎠ ⎟
1
m
60 5 =
. ft 18.4
3.281 ft
m
The time to reach home plate is
t
= Δ x
= 18.4 m
=
v45.1 m s
0
x
0.408 s
In this time interval, the vertical displacement is
1 2 2 2
2
Δy = v t + at = 0
+ (− )( ) = − 0
y y 1
2
9.80 m s 0.408 s 0.817 m
Thus, the ball drops vertically 0.817 m (3.281 ft 1 m) = 2.68 ft .

90 Chapter 3
3.23 (a) With the origin chosen at point O as shown in Figure P3.23, the coordinates of the original
position of the stone are x0 = 0 and y0 = +50.0 m .
(b) The components of the initial velocity of the stone are v v 0 0 18 0 0 x y = + . m s and = .
(c) The components of the stone’s velocity during its fl ight are given as functions of time by
v v v x x x x = + a t = + ( )t = 0 18.0 m s 0 or 18.0 m s
and
v v v y y y x = + a t = + (−g)t = −( )t 0 0 or 9.80 m s2
(d) The coordinates of the stone during its fl ight are
1 2 2
2
x = x + t + at = 0 +( 180
) t + ( ) t 0 0
x x 1
2
v . m s 0 or x = (18.0 m s)t
and
1 2 2
2
y = y + t + at = 50 0 + ( 0
) t + (− g ) t 0 0
y y 1
2
v . m or y = 50.0 m − (4.90 m s2 )t 2
1 2
2
(e) We fi nd the time of fall from Δy t at y y = v + 0
with v0 0 y = :
t
= ( ) =
Δ .
y
a
(− )
−
=
2 2 500
9 .
80
m
m s
3.19 s 2
(f ) At impact, v v x x = = 0 18 0. m s, and the vertical component is
v v y y y = + a t = + (− )( ) = − 0 0 9.80 m s2 3.19 s 31.3 m s
Thus,
v = v + v = ( ) + (− ) = x y
2 2 2 2 18.0 m s 31.3 m s 36.1 m s
and
θ =
⎛
1 1 . 31 3
⎝ ⎜
⎞
⎠ ⎟
= − ⎛⎝
⎞⎠
.
.
tan− tan− = − °
18 0
60 1
v
v
y
x
or v
= 36.1 m s at 60.1° below the horizontal .
3.24 The constant horizontal speed of the falcon is
vx =
⎛
200 =
⎝ ⎜
⎞
⎠ ⎟
0 447
1
89 4
mi
h
m s
mi h
m s
.
.
The time required to travel 100 m horizontally is
t
= Δ x
= =
v
x
100
1 12
m
89.4 m s
. s
The vertical displacement during this time is
1 2 2
2
Δy = v t + at = 0
+ (− )( ) = − 0
y y 1
2
9.80 m s2 1.12 s 6.13 m
or the falcon has a vertical fall of 6.13 m .

3.25 At the maximum height vy = 0, and the time to reach this height is found from
vy = v y + ayt 0 as t
−
v v v v 0 0 0 0
=
y y
y y a g g
y
=
−
−
=
The vertical displacement that has occurred during this time is
⎛ +
Δy t t y
( ) = ( ) = y y y
⎝ ⎜
⎞
⎠ ⎟
=
⎛ +
⎝ ⎜
⎞
⎠ ⎟
v
max av
v v v 0 0
2
0
2
2
2
v v 0 0
y y
g g
⎛
⎝ ⎜
⎞
⎠ ⎟
=
Thus, if (Δy) ( . max = 12 ft 1 m 3.281 ft ) = 3 7 m, then
v0 2 29 80 3 7 8 5 y = g(Δy) = ( )( ) = max
. m s2 . m . m s
and if the angle of projection is θ = 45°, the launch speed is
v
v
0
0 8 5
= y
= = 12
°
45
sin
.
θ sin
m s
m s
3.26 (a) When a projectile is launched with speed v0 at angle θ0 above the horizontal, the initial
velocity components are v v v v 0 x 0 0 0 y 0 0 = cosθ and = sinθ . Neglecting air resistance, the
vertical velocity when the projectile returns to the level from which it was launched (in this
case, the ground) will be v v y y = − 0 . From this information, the total time of fl ight is found
from v v y y y = + a t 0 to be
t
−
v v v v v 0 0 0 0 2
t yf y
a g g
y
y y y
= =
= or
total t − −
−
otal = 20 0 v sinθ
g
Since the horizontal velocity of a projectile with no air resistance is constant, the horizontal
distance it will travel in this time (i.e., its range) is given by
θ
= =( )⎛
x g g 2 2
v sin
v
v v =
R t
0 2 ( ) = v ( )
θ s
⎝ ⎜
⎞
⎠ ⎟
0 0 0
0 0 0
2 total cos
in cos
sin
θ θ
θ
0 0
2
0
g
Thus, if the projectile is to have a range of R = 81 1 . m when launched at an angle of
θ0 = 45.0°, the required initial speed is
v0
m ms2
= Rg
0 2
81 1 9 80
90 0
= ( ) =
( )( )
( °)
sin
. .
θ sin .
28 2. m s
(b) With v0 = 28.2 m s and θ0 = 45.0° the total time of fl ight (as found above) will be
t
2 v sin 2(28 . 2 ) sin (45 .
0°)
= 0 0 = g total
m s
m
.
9 80
θ
s
s 2 = 4.07
(c) Note that at θ0 = 45.0°, sin (2 ) 1 0 θ = and that sin (2 ) 0 θ will decrease as θ0 is increased above
this optimum launch angle. Thus, if the range is to be kept constant while the
launch angle is increased above 45.0°, we see from v0 0 = Rg sin(2θ ) that
the required initial velocity will increase .
Observe that for θ0 90°, the function sinθ0 increases as θ0 is increased. Thus, increas-ing
the launch angle above 45.0° while keeping the range constant means that both
v0 0 and sinθ will increase. Considering the expression for ttotal given above, we see that
the total time of flight will increase .

92 Chapter 3
3.27 When Δy Δy y = ( ) , = max v 0.
Thus, v v y y y = + a t 0 yields 0 300 0 = v sin . ° − gt, or
t
g
=
v ° 0 sin3.00
1 2
2
The vertical displacement is Δy t at y y = v + 0
. At the maximum height, this becomes
3 00 1
( Δy
) = ( °) − g
max
g
⎛
⎝ ⎜
⎞
⎠ ⎟
°
v
v v
0
3 00 0 0
2
sin .
° 2 2 °
sin . sin3.00 sin 3.00
2
2
0
g g
⎛
⎝ ⎜
⎞
⎠ ⎟
= v
If (Δy) . max = 0 330 m, the initial speed is
v0
= g Δy ( )( ) max
2
3 00
2 9 80 0 330
=
( )
°
m s2 m
sin .
. .
sin .
.
3 00
48 6
°
= m s
Note that it was unnecessary to use the horizontal distance of 12.6 m in this solution.
3.28 (a) With the origin at ground level directly below the window, the original coordinates of the
ball are (x, y) = (0, y ) 0 .
(b) v v 0 0 0 8 00 20 0 7 52 x= cosθ = ( . m s)cos(− . °) = + . m s
v v 0 0 0 8 00 20 0 2 74 y= sinθ = ( . m s)sin(− . °) = − . m s
1 2 2
2
(c) x = x + t + at = 0 +( 752
) t + ( ) t 0 0
x x 1
2
v . m s 0 or x = (7.52 m s)t
y = y + t + at 2
= y + (− ) t + − 0 0
y y 0
1
2
2 74
1
2
v . m s ( 9.80 m s2 )t 2
or y = y − ( )t − ( )t 0
2.74 m s 4.90 m s2 2
(d) Since the ball hits the ground at t = 3.00 s, the x-coordinate at the landing site is
m s s landing s x xt = =( )( ) = =3 00 7 52 3 00 22 6 . . . . m
(e) Since y = 0 when the ball reaches the ground at t = 3.00 s, the result of (c) above gives
m
s
y y t t
t
0
2 2 74 4 90 = +⎛⎝ ⎜
⎞⎠ ⎟+⎛⎝ ⎜
⎞⎠ ⎟
⎡
⎣ ⎢
⎤
⎦ ⎥
. .
m
s2
=
= +⎛⎝ ⎜
⎞⎠ ⎟
( ) +⎛⎝
3 00
0 2 74 3 00 4 90
.
. . .
s
2
m
s
s
m
⎜ s
⎞⎠ ⎟
(3 00 )2 . s
or y0 = 52.3 m .
(f ) When the ball has a vertical displacement of Δy = −10 0 . m, it will be moving downward
= 2 + 2 (Δ ) as
with a velocity given by v v y y y 2 a y
0
v v y y y = − + a ( y) = − (− ) + (− ) − 0
2 2 2 Δ 2.74 m s 2 9.80 m s2 ( 10.0 m) = −14.3 m s
The elapsed time at this point is then
t
y y
a
y
=
−
=
− − (− )
−
=
v v0 14 3 2 74
9 80
1 1
. .
.
.
m s m s
m s2 8 s

3.29 We choose our origin at the initial position of the projectile. After 3.00 s, it is at ground level, so
the vertical displacement is Δy = −H.
1 2
2
To fi nd H, we use Δy t at y y = v + 0
, which becomes
H 15 25° (3 0 ) + (− )
− = ( ) ⎡⎣
⎤⎦
1
2
m s sin . s 9.80 m s2 3.0 2 ( s) , or H = 25 m
3.30 The initial velocity components of the projectile are
v0 300 55 0 172 x = ( m s)cos . ° = m s and v0 300 55 0 246 y = ( m s)sin . ° = m s
while the constant acceleration components are
ax = 0 and a g y = − = −9.80 m s2
The coordinates of where the shell strikes the mountain at t = 42 0 . s are
1 2 3
2
x t at x x = v + = ( )( ) + = × 0
172 m s 42.0 s 0 7.22 10 m = 7.22 km
and
1 2
2
y t at y y = +
v0
= ( 246 )( 42 0
) + (−
1
2
m s . s 9.80 m s2 )(42 0 ) = 1 69 ×10 = 1 69 . s 2 . 3 m . km
3.31 The speed of the car when it reaches the edge of the cliff is
v = v + ( ) = + ( )( ) = 0
2 2a Δx 0 2 4.00 m s2 50.0 m 20.0 m s
Now, consider the projectile phase of the car’s motion. The vertical velocity of the car as it
reaches the water is
2 Δ 20.0 m s sin 24.0 2(−9.80 m s2 )(−30.0 m)
v v y y y a y = − + ( ) = − −( ) ° ⎡⎣
2 2
⎤⎦
+ 0
or
vy = −25 6. m s
(b) The time of fl ight is
t
v v25.6 m s 20.0 m s sin 24.0
y 0 y
a
y
=
−
=
− − − ( ) ⎡⎣
°
⎤⎦
−
9 80
1 78
.
.
m s
s 2 =
(a) The horizontal displacement of the car during this time is
Δx tx = = ( ) ⎡⎣
v ° ( ) = 0 20.0 m s cos24.0 1.78 s 32.5 m
⎤⎦

94 Chapter 3
3.32 The components of the initial velocity are
v0 x = (40.0 m s)cos30.0° = 34.6 m s
and
v0 40 0 30 0 20 0 y = ( . m s)sin . ° = . m s
The time for the water to reach the building is
t
= Δ x
= 50 0
=
v0
x
1 44
.
.
m
34.6 m
s
The height of the water at this time is
1 2
2
Δy = v t + at = ( 20 0 )( 1 44
) + − 0
y y 1
2
. m s . s ( 9.80 m s2 )(1 44 s) = 18 7 m 2 . .
3.33 (a) At the highest point of the trajectory, the projectile is moving horizontally with velocity
components of vy = 0 and
v v v x x = = =( ) ° = 0 0 cosθ 60.0 m/s cos30.0 52.0 m s
(b) The horizontal displacement is Δx tx = v = ( )( ) = 0 52.0 m s 4.00 s 208 m and, from
Δy t a t y = (v sin ) + 0
θ 2, the vertical displacement is
1
2
1
2
. m s sin . . s 9.80 m s2 )(4 00 ) = 41 6 2 . s . m
Δy = (60 0 )( 30 0°) (4 00 ) + (−
The straight line distance is
d = (Δx) + (Δy) = ( ) + ( ) = 2 2 2 2 208 m 41.6 m 212 m
3.34 The horizontal kick gives zero initial vertical velocity to the rock. Then, from Δy = v t + 1
at 0
y 2
y 2,
the time of fl ight is
t
= ( ) =
y
ay
(− )
−
=
2 2 400
9 80
8 16
Δ .
.
.
m
m s
s 2
The extra time Δt = 3.00 s − 8.16 s = 0.143 s is the time required for the sound to travel in a
straight line back to the player. The distance the sound travels is
d = (Δx) + (Δy) = Δt = ( )( ) = 2 2 v 343 0 143 49 sound m s . s .0 m
where Δx represents the horizontal displacement of the rock when it hits the water. Thus,
Δx = d2 − (Δy)2 = ( )2 − (− )2 = 49.0 m 40.0 m 28.3 m
The initial velocity given the ball must have been
v v 0 0
Δ .
28 3
x
t
= = = = 9 .
91 x
m
8.16 s
m s

3.35 (a) The jet moves at 3.00 × 102 mi h due east relative to the air. Choosing a coordinate system
with the positive x-direction eastward and the positive y-direction northward, the compo-nents
of this velocity are
v v JA JA ( ) = × mi h and ( ) = x y 3.00 102 0
(b) The velocity of the air relative to Earth is 1.00 × 102 mi h at 30.0° north of east. Using the
coordinate system adopted in (a) above, the components of this velocity are
v v AE AE ( ) = = ( × mi h)cos30.0°= 86. x cosθ 1.00 102 6 mi h
and
v v AE AE ( ) = = ( × mi h)sin30.0° = 50. y sinθ 1.00 102 0 mi h
(c) Carefully observe the pattern of the subscripts in Equation 3.16 of the textbook. There, two
objects (cars A and B) both move relative to a third object (Earth, E). The velocity of object
A relative to object B is given in terms of the velocities of these objects relative to E as
v v v AB AE BE = − . In the present case, we have two objects, a jet (J) and the air (A), both
moving relative to a third object, Earth (E). Using the same pattern of subscripts as that in
Equation 3.16, the velocity of the jet relative to the air is given by
v v v JA JE AE = −
(d) From the expression for v
JA found in (c) above, the velocity of the jet relative to the ground
is v v v JE JA AE = + . Its components are then
v v v JE JA AE ( ) = ( ) + ( ) = × mi h + 86.6 m x x x 3.00 102 i h = 3.87 ×102 mi h
and
v v v JE JA AE ( ) = ( ) + ( ) = + 50.0 mi h = mi h y y y 0 50 0 .
This gives the magnitude and direction of the jet’s motion relative to Earth as
v v v JE JE JE = + = ( × mi h) + mi h x y
2 2 2 2 3.87 10 (50.0 ) = × 2 3.90 102 mi h
and
θ =
( )
( )
⎛
⎛
1 1 50 0
tan− = tan−
⎝ ⎜
⎞
⎠ ⎟
.
.
3

v
v
JE
JE
y mi h
7 37 × 2
⎝ ⎜
x 87 10
⎞
⎠ ⎟
= °
mi h
.
Therefore, v
JE = 3.90 ×102 mi h at 7.37° north of east .

96 Chapter 3
3.36 We use the following notation:
v
BS = velocity of boat relative to the shore
v
BW = velocity of boat relative to the water,
and v
WS = velocity of water relative to the shore.
If we take downstream as the positive direction, then v
WS = +1.5 m s for both parts of the trip.
Also, v
BW = +10 m s while going downstream and v
BW = −10 m s for the upstream part of
the trip.
The velocity of the boat relative to the boat relative to the water is v v v BW BS WS = − , so the
velocity of the boat relative to the shore is v v v BS BW WS = + .
While going downstream, v
BS = 10 m s +1.5 m s and the time to go 300 m downstream is
t
d
down
300
( + ) = v
= = s
BS
m
10 1.5 m s
26
When going upstream, v
BS = −10 m s +1.5 m s = −8.5 m s and the time required to move 300 m
upstream is
t
d
up
300
s = = = v
BS
m
8.5 m s
35
The time for the round trip is t = t + t = ( + ) = down up 26 35 s 61 s .
3.37 Prior to the leap, the salmon swims upstream through water fl owing at speed
v
WE = 1.50 m s relative to Earth. The fi sh swims at v
FW = 6.26 m s relative
to the water in such a direction to make its velocity relative to Earth, v
FE,
vertical. Since v v v FE FW WE = + , as shown in the diagram at the right, we fi nd
that
θ =
⎛
− 1 − 11 50
⎞ cos cos
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
.
.
6 26

v
v
WE
FW
m s
m s⎠ ⎟= 76.1°
and the vertical velocity of the fi sh as it leaves the water is
v0 6 26 76 1 6 08 y= v = v = ( ) ° = FE FW sinθ . m s sin . . m s
The height of the salmon above the water at the top of its leap (that is, when vy = 0)
is given by
Δy
2 2
y y
a
y
=
−
=
− ( )
(− ) =
v2 v
0
2
0 608
2 980
m s
1 8
8 m
m s2 .
.
.
→

3.38 (a) The velocity of the boat relative to the water is
v v v BW BS WS = −
where v
BS is the velocity of the boat relative to shore, and v
WS is
the velocity of the water relative to shore. Thus, we may write
v v v BS BW WS = + . Note, in the vector diagram at the right, that
these vectors form a right triangle. The Pythagorean theorem then
gives the magnitude of the resultant v
BS as
v v v BS BW WS = + = ( m s) + ( m s) = 2 2 2 2 10.0 1.50 10.1 m s
also,
θ =
⎛
1 11 50
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝
tan− tan−
.
.
10 0

v
v
WS
BW
m s
⎜ m s
⎞
⎠ ⎟
= 8.53°
so v
BS = 10.1 m s at 8.53° east of north .
(b) The boat has a constant northward velocity of v
VBW VBS
BW = 10.0 m s, and must travel a distance
d = 300 m northward to reach the opposite shore. The time required will be
t
d = = = v
BW
m
m s
s
300
10 0
30 0
.
.
During this time, the boat drifts downstream at a constant rate of vWS = 1.50 m s.
The distance it drifts downstream during the crossing is
drift t = ⋅ = ( )( ) = v
WS 1.50 m s 30.0 s 45.0 m
3.39 v
BW = velocity of boat relative to the water
v
WS = velocity of water relative to the shore
v
BS = velocity of boat relative to the shore
v v v BS BW WS = + (as shown in the diagram)
The northward (that is, cross-stream) component of v
BS is
v
BS north BW ( ) = (v )sin62.5° + 0 = (3.30 mi h)sin62.5° + 0 = 2.93 mi h
The time required to cross the stream is then
t= = 0 505
0 173
.
.
mi
2.93 mi h
h
The eastward (that is, downstream) component of v
BS is
v
BS east BW WS ( ) = −(v )cos62.5° + v
= −(3.30 mi h)cos62.5° + 1.25 mi h = − 0.274 mi h
Since the last result is negative, it is seen that the boat moves upstream as it crosses the river.
The distance it moves upstream is
d t = ( ) = ( )( ) = × − v
BS east 0.274 mi h 0.173 h 4.72 10 2 5 280
ft
( mi )⎛⎝
249
ft 1 mi
⎞⎠
=
north
east
VWS

98 Chapter 3
3.40 If the salmon (a projectile) is to have vy = 0 when Δy = +1.50 m, the required initial velocity in
= 2 + 2 Δ as
the vertical direction is given by v v y y y 2 a y
0
2 2 0 2 9 80 1 50 y y y = + − a Δy = − (− . m s2 )(+ . m) = 5.42 m s
v v 0
The elapsed time for the upward fl ight will be
Δt
y y
a
y
=
−
= −
−
=
v v0 0 542
9 80
0 553
.
.
.
m s
m s
s 2
If the horizontal displacement at this time is to be Δx = +1.00 m, the required constant horizontal
component of the salmon’s velocity must be
v0
= Δ = 1 00
=
Δ
x
x
t
. m
0.553 s
1.81m s
The speed with which the salmon must leave the water is then
= 2
+ 2 = (1 . 81 m s ) 2 + (5 . 42 m s ) 2 = 5 . 72 m s
x y v v v 0 0
0
Yes , since v0 6.26 m s the salmon is capable of making this jump.
3.41 (a) Both the student (S) and the water (W) move relative to Earth (E). The velocity of the
student relative to the water is given by v v v SW SE WE = − , where v v SE WE and are the
velocities of the student relative to Earth and the water relative to Earth, respectively. If we
choose downstream as the positive direction, then v
WE = + 0.500 m s, v
SW = −1.20 m s
when the student is going up stream, and v
SW = +1.20 m s when the student moves
downstream.
The velocity of the student relative to Earth for each leg of the trip is
v v v SE upstream WE SW upstream ( ) = + ( ) = 0.500 m s + (−1.20 m s) = − 0.700 m s
and
v v v SE downstream WE SW downstream ( ) = + ( ) = 0.500 m s + (+1.20 m s) = +1.70 m s
The distance (measured relative to Earth) for each leg of the trip is d = 1.00 km =
1.00 × 103 m. The times required for each of the two legs are
t
d
upstream
SE upstream
. 3
.
m
m s
= = ×
v
1 00 10
0 700
= 1.43 ×103 s
and
t
= d
= ×
m downstream
SE downstream
v
. 3
. m s
1 00 10
1 70
= 5.88 ×10s s
so the time for the total trip is
t t t total upstream downstream = + = 1.43 ×103 s + 5.88 ×10s s = 2.02 ×103 s
continued on next page

(b) If the water is still v
( WE = 0), the speed of the student relative to Earth would have been the
same for each leg of the trip, v v v SE SE upstream SE downstream = = = 1.20 m s. In this case, the time
for each leg, and the total time would have been
t
v
= d
= × = × s leg
SE
m
1 00 10
1 20
m s
8 33 10
3
. 3
.
. and s total leg t = 2t = 1.67 ×103
(c)
The time savings going downstream with the current is always less than the extra time
required to go the same distance against the current.
3.42 (a) The speed of the student relative to shore is v v v up= − s while swimming upstream and
v v v down= + s while swimming downstream. The time required to travel distance d upstream
is then
t
d d
s
up
= =
v v − v
up
(b) The time required to swim the same distance d downstream is
t
d d
s
down
= =
v v + v
down
(c) The total time for the trip is therefore
v v v v
v vs s s s
= + = s s
t t t
d d d d
a
s s
−
+
+
=
( + )+ ( − )
− up down v v v v
v
d d
( )( + ) =
−
=
v v v v
−
v
v v
2 2
2 2 1 2 2
(d) In still water, vs = 0 and the time for the complete trip is seen to be
= =
t t
2
1 0
2
d d
b a
= v =
s
−
v
0 v2 v
(e) Note that t
v
v
b= = 2 2
d d
v 2
and that t
2
d
2 2
a
s
=
−
v
v v
.
Thus, when there is a current (v ) s 0 , it is always true that ta tb .

100 Chapter 3
3.43 (a) The bomb starts its fall with v0 y = 0 and v0 = v = 275 m s x plane . Choosing the origin at the
location of the plane when the bomb is released and upward as positive, the y coordinate of
the bomb at ground level is y = −h = −3.00 × 103 m. The time required for the bomb to fall
is given by y = y + v t + 1
at 2 as
0 0
y 2
y 1
2
−3 00 ×10 = 0 + 0 + (− )
. 3 m 9.80 m s2 2
( × ) =
2 3 00 10
9 80
= s
fall t or tfall 2
m
m s
24 7
. 3
.
.
With ax = 0, the horizontal distance the bomb travels during this time is
d = v t = ( )( ) = × = 0
275 24 7 6 80 103 6 80 x fall m s . s . m . km
(b) While the bomb is falling, the plane travels in the same horizontal direction with the
same constant horizontal speed, v v v x x = = 0 plane, as the bomb. Thus, the plane remains
directly above the bomb as the bomb falls to the ground. When impact occurs, the plane is
directly over the impact point, at an altitude of 3.00 km.
(c) The angle, measured in the forward direction from the vertical, at which the bombsight
must have been set is
θ = ⎛⎝ ⎜
⎞⎠ ⎟
= ⎛⎝ ⎜
⎞⎠ ⎟
.
.
1 1 6 80
tan− tan− =
3 00
6
d
h
km
km
6 2 . °
3.44 (a) The time required for the woman, traveling at constant speed v1 relative to the ground, to
travel distance L relative to the ground is t L woman = v1 .
(b) With both the walkway (W) and the man (M) moving relative to Earth (E), we know that
the velocity of the man relative to the moving walkway is v v v MW ME WE = − . His velocity
relative to Earth is then v v v ME MW WE = + . Since all of these velocities are in the same
direction, his speed relative to Earth is v v v ME MW WE = + = v + v 2 1. The time required for
the man to travel distance L relative to the ground is then
t
L L
man
= =
+ v
ME
v v 1 2
3.45 Choose the positive direction to be the direction of each car’s motion relative to Earth.
The velocity of the faster car relative to the slower car is given by v v v FS FE SE = − , where
v
FE = + 60.0 km h is the velocity of the faster car relative to Earth, and v
SE = 40.0 km h is the
velocity of the slower car relative to Earth.
Thus, v
FS = +60.0 km h − 40.0 km h = + 20.0 km h and the time required for the faster car to
move 100 m (0.100 km) closer to the slower car is
t
km
= d = = × −
v20.0 km h
FS
h
3600 s
1
0 100
5 00 10 3 .
.
s ⎛⎝
h
⎞⎠
= 18 0 .
3.46 The vertical displacement from the launch
point (top of the building) to the top of
the arc may be found from v 2 = v 2 + 2 a Δ
y
y y y 0
with vy = 0 at the top of the arc. This yields
Δy
2 2
y y
a
y
=
−
=
− ( )
(− ) = +
v2 v
0
2
0 120
2 980
m s
7
35 m
m s2 .
.
.
and Δy = y − y max 0 gives
y y y y max = + = + . 0 0 Δ 7 35 m

(a) If the origin is chosen at the top of the building, then y0 = 0 and y = 7 35 m max . .
Thus, the maximum height above the ground is
h y max max = 50.0 m + = 50.0 m + 7.35 m = 57.4 m
The elapsed time from the point of release to the top of the arc is found from v v y y y = + a t 0 as
t
y y
a
y
=
−
= −
−
=
v v0 0 120
9 80
1 22
.
.
.
m s
m s
s 2
(b) If the origin is chosen at the base of the building (ground level), then y0 = +50.0 m and
h y max max = , giving
h y y max = + = . + . = . 0 Δ 50 0 m 7 35 m 57 4 m
The calculation for the time required to reach maximum height is exactly the same as that
given above. Thus, t = 1.22 s .
3.47 (a) The known parameters for this jump are: θ0 = −10.0°, Δ x = 108 m, Δy = −55.0 m, ax = 0,
and a g y = − = −9.80 m s.
Since ax = 0, the horizontal displacement is Δx t t x = v = (v ) 0 0 0 cosθ where t is the total time
of the fl ight. Thus, t = Δx (v ) 0 0 cosθ .
The vertical displacement during the fl ight is given by
1 2
2 2
Δy t at t
gt
= v + 2
= (v sinθ
) − 0
y y 0 0
or
Δ Δ Δ
y
x g x = ( )⎛
⎝ ⎜
⎞
⎠ ⎟
−
⎛
⎝ ⎜
⎞
v
v v 0 0
0 0 0 0 2
sin
cos cos
θ
= ( ) − ⎡ ( )
1 Δ
θ θ⎠ ⎟
⎣ ⎢
⎤
⎦ ⎥
2
0
2
2 2
2
0 0
Δ
x
g x
tan
cos
θ
θ v
Thus,
Δ − ( Δ
) ⎡⎣
g x Δ
y x
⎤⎦
= − ⎡ ( )
⎣ ⎢
⎤
⎦ ⎥
tan
1
v
2 2
cos
θ
2
0
2
θ 0 0
or
v0
g x 2
−
0
2
0 2
9 80
=
− ( )
− ( ) ⎡⎣
⎤⎦
=
y x
Δ
Δ Δ tan cos
.
θ θ
( m s2 )( )
− − ( ) − ( ) ° ⎡⎣
⎤⎦
108
2 55 0 108 10 0
2 m
. m m tan . cos2 (−10.0°)
yielding
v0
= − 1 .
143 ×
105
= 40 5 69 75
−
.
.
m s
m
m s
3 2
(b) Rather than falling like a rock, the skier glides through the air much like a bird, prolonging
the jump.

102 Chapter 3
3.48 The cup leaves the counter with initial velocity components of (v v , v ) 0 0 0 x i y = = and has
acceleration components of (a , a g) x y = 0 = − while in fl ight.
(a) Applying Δy t at y y = v + 0
2 2 from when the cup leaves the counter until it reaches the fl oor
gives
(− ) h
− = +
g
0 t
2
2
so the time of the fall is
t
= 2
h
g
(b) If the cup travels a horizontal distance d while falling to the fl oor, Δx tx = v0 gives
v v i x
Δ
x
t
d
h g
= = = 0 2
or vi d
g
h
=
2
(c) The components of the cup’s velocity just before hitting the fl oor are
v v v x x i d
g
h
= = = 0 2
and v v y y y a t g
2
h
g
= + = 0
− = − 2
gh 0 Thus, the total speed at this point is
2
2
d g
h
v = v + v = + x y
2 2 gh
2
(d) The direction of the cup’s motion just before it hits the fl oor is
θ =
⎛
tan−1 tan−1 = tan−1 2
⎝ ⎜
⎞
⎠ ⎟
=
⎛ −
⎝ ⎜
⎞
⎠ ⎟
2
v
v
y
x
gh
d g h
− ⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜⎜
⎞
⎠ ⎟⎟
= − ⎛⎝
⎞⎠
1 −
2
2 1 2
d
gh
h
g
h
d
tan
3.49 AL = v t = ( )( ) = 1 90.0 km h 2.50 h 225 km
BD = AD − AB = AL cos 40.0° − 80.0 km = 92.4 km
From the triangle BCD,
= ( ) + ( )
= ( ) + ( °)
BL BD DL
AL
2 2
2 2
92.4 km sin 40.0 =172 km
Since car 2 travels this distance in 2.50 h, its constant speed is
v2
172
= = 68 6 km
2.50 h
. km h

3.50 After leaving the ledge, the water has a constant horizontal component of velocity.
vx = v0 x = 1.50 m s2
Thus, when the speed of the water is v = 3.00 m s, the vertical component of its velocity will be
v v v y x = − 2 − 2 = − ( )2 − ( )2 = − 3.00 m s 1.50 m s 2.60 m s
The vertical displacement of the water at this point is
Δy
2 2
y y
a
y
=
−
=
(− ) −
(− ) = −
v2 v
0
2
2 60 0
2 980
0
.
.
m s
m s2 .344 m
or the water is 0.344 m below the ledge.
If its speed leaving the water is 6.26 m s, the maximum vertical leap of the salmon is
Δy
2 2 v .
a
y
y
leap
m s
2 9.80 m s
=
−
=
− ( )
(− ) =
0
2
0 626
2 0
.00 m
Therefore, the maximum height waterfall the salmon can clear is
h y max = Δ + . = . leap 0 344 m 2 34 m
3.51 The distance, s, moved in the fi rst 3.00 seconds is given by
1 2
2
s = v t + at = ( 100 )( 3 00
) + ( ) 0
1
2
m s . s 30.0 m s2 3.00 435 2 ( s) = m
Choosing the origin at the point where the rocket was launched, the coordinates of the rocket at
the end of powered fl ight are:
x s 1 = (cos53.0°) = 262 m and y s m 1 = (sin 53.0°) = 347
The speed of the rocket at the end of powered fl ight is
v v 1 0 = + at = 100 m s + (30.0 m s2 )(3.00 s) = 190 m s
so the initial velocity components for the free-fall phase of the fl ight are
v v 0 1 53 0 114 x= cos . ° = m s and v v 0 1 53 0 152 y= sin . ° = m s
(a) When the rocket is at maximum altitude, vy = 0. The rise time during the free-fall phase can
be found from v v y y y = + a t 0 as
t
−
15 5 0 v
= −
−
rise 2
a
y
y
m
m s
= =
s
0 0 152
9 .
80
.
The vertical displacement occurring during this time is
⎛ +
( ) v v0
Δy t y y =
⎝ ⎜
⎞
⎠ ⎟
= + ⎛⎝
⎞⎠
2
0 152
1 rise
2
m s
5.5 s = 1.17 ×103 m
The maximum altitude reached is then
H = y + y = + × = × 1
Δ 347 m 1.17 103 m 1.52 103 m

104 Chapter 3
(b) After reaching the top of the arc, the rocket falls 1.52 × 103 m to the ground, starting with
zero vertical velocity (v = 0 ) . The time for this fall is found from Δy = v t + 1
at 0 y 0
y 2
y 2 as
t
y
ay
fall 2
m
9.80 m s
= ( ) =
(− × )
−
=
2 2 1 52 10
17 6
3 Δ .
. s
The total time of fl ight is
t = t + t + t = ( + + ) = powered rise fall 3.00 15.5 17.6 s 36.1 s
(c) The free-fall phase of the fl ight lasts for
t t t 2 = + = (15 5 +17 6) = 33 1 rise fall . . s . s
The horizontal displacement occurring during this time is
Δx tx = v = ( )( ) = × 0 2
114 m s 33.1 s 3.78 103 m
and the full horizontal range is
R = x + x = + × = × 1
Δ 262 m 3.78 103 m 4.05 103 m
3.52 Taking downstream as the positive direction, the velocity of the water relative to shore is v
WS WS = +v , where vWS is the speed of the fl owing water. Also, if vCW is the common speed of the
two canoes relative to the water, their velocities relative to the water are
CW downstream CW ( ) = +v and v
v
CW upstream CW ( ) = −v

CW CS WS = − .
The velocity of a canoe relative to the water can also be expressed as v v v
Applying this to the canoe moving downstream gives
+ v = + − v CW WS 2.9 m s [1]
and for the canoe going upstream
−v = − − v CW WS 1.2 m s [2]
(a) Adding equations [1] and [2] gives
2v 1 7 WS = . m s, so vWS = 0.85 m s
(b) Subtracting [2] from [1] yields
2v 4 1 CW = . m s, or vCW = 2.1 m s

3.53 The time of fl ight is found from Δy = v t + 1
t 0
y 2
ay2 with Δy = 0, as
t
2 0 v
y =
g
This gives the range as
v v
x y = v =
R t
g x
0
0 0 2
On Earth this becomes
R
g
x y
Earth
Earth
=
2 0 0 v v
and on the moon,
R
g
x y
Moon
Moon
=
2 0 0 v v
Dividing RMoon by REarth, we fi nd R g g R Moon Earth Moon Earth =( ) . With g g Moon Earth = (1 6) , this gives
R R Moon Earth = 6 = 6(3.0 m) = 18 m .
Similarly,
R
g
g
= Earth
7
R Mars
Mars
Earth
m
0.38
⎛
⎝ ⎜
⎞
⎠ ⎟
= = 3 0
9
.
. m .
3.54 The time to reach the opposite side is
t
= Δ x
=
v v 0 x
0
10
m
cos 15
°
When the motorcycle returns to the original level, the vertical displacement is Δy = 0. Using this
in the relation Δy = v t + 1
at 2 gives a second relation between the takeoff speed and the time
0
y 2
y of fl ight as
1
2 0
= (v sin °)t + (−g)t 2 or v0 2 15
0 15
=
⎛
⎞
⎟
⎠ ⎜⎝ g
t
sin °
Substituting the time found earlier into this result yields
v
v 0
0 2 15
10
m
15
=
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
g
sin ° cos
°
or v0
= 14 ( )( )
9 80 10
2 15 15
( °) ( °)
=
.
sin cos
m s m
m s
2

106 Chapter 3
3.55 (a) The time for the ball to reach the fence is
t
= Δ x
= 130
m =
159
v v 35
°
v 0 x
0 0
m
cos
At this time, the ball must be Δy = 21m − 1.0 m = 20 m above its launch position, so
Δy = v t + 1
at 2 gives
0
y 2
y = ( °) m s2
v − ( )
20 35
159
sin .
4 90
⎛
2 159
0
0
m
m
⎛
⎝ ⎜
⎞
⎠ ⎟
v
m
v0
⎝ ⎜
⎞
⎠ ⎟
or
( )( )
( 159 ) 35 °− 20
= 4 90 m s2 159 m
2
2 m m
0
sin
.
v
From which, v0
= 42 ( )( )
4 90 2
159
2 159 35 20
m s m
m m
( ) °−
=
.
sin
m s
(b) From above, t= = = 159 159
3 8
0
m m
42 m s
s
v
. .
(c) When the ball reaches the wall (at t = 3 8 . s),
v v x x = =( ) ° = 0 42 m s cos35 34 m s
v v y y y = + a t = ( ) ° − ( )( ) = 0 42 m s sin35 9.80 m s2 3.8 s −13 m s
v = v + v = ( ) + (− ) = x y
2 2 2 2 34 m s 13 m s 37 m s
3.56 We shall fi rst fi nd the initial velocity of the ball thrown vertically upward, recognizing that it
takes descend from its maximum height as was required to reach this height. Thus, the elapsed
time when it reaches maximum height is t = 1.50 s. Also, at this time, vy = 0, and, v v y y y = + a t 0
gives
0 9 80 1 50 0 = v − ( )( ) y . m s2 . s or v0 14 7 y = . m s
In order for the second ball to reach the same vertical height as the fi rst, the second must have the
same initial vertical velocity as the fi rst. Thus, we fi nd v0 as
v
v
0
0
30 0
14 7
0 500
= = y
= 29 4
°
sin .
.
.
.
m s
m s
3.57 The time of fl ight of the ball is given by Δy = v t + 1
at 0
y 2
y 2, with Δy = 0, as
m s sin ° t + (− . m s2 )t
0 20 30
1
2
9 80 2 = ( ) ⎡⎣
⎤⎦
or t = 2.0 s.
The horizontal distance the football moves in this time is
Δx tx = = ( ) ⎡⎣
v ° ( ) = 0 20 m s cos30 2.0 s 35 m
⎤⎦
Therefore, the receiver must run a distance of Δx = 35 m − 20 m = 15 m away from the
quarterback, in the direction the ball was thrown to catch the ball. He has a time of 2.0 s to do
this, so the required speed is
v = = = Δx
t
15
7 5
m
2.0 s
. m s

3.58 The horizontal component of the initial velocity is v0 x = v0 cos40° = 0.766 v0, and the time
required for the ball to move 10.0 m horizontally is
t
= Δ x
= 10 0
=
v 0 766
v v 0 x
0
. 13 1
.
m . m
0
At this time, the vertical displacement of the ball must be
Δy = y − y = − = 0 3.05 m 2.00 m 1.05 m
Thus, Δy = v t + 1
at 0
y 2
y 2 becomes
= ( v °) + (− m s2 )
1 05 40 0
13 1 1
2
9 80
( 13
m)
0
0
. sin .
.
m
m .
v
.1 2
2
0
v
or
1 05 8 39
835
m3 s2
2 . m . m
0
= −
v
which yields
v0
835
8 39 1 05
= m s
m m
= 10 7
−
m s
3 2
. .
.
3.59 Choose an origin where the projectile leaves the gun and
let the y-coordinates of the projectile and the target at time t
be labeled y y p T and , respectively.
Then, ( ) ( sin ) Δy y t g
t p p = − 0 = −
v θ 2, and
2 0 0
g
t T T = − = 0 −
(Δy) y h
2
g
t T = −
2 or y h
2
2
The time when the projectile will have the same
x-coordinate as the target is
t
= Δ =
v v 0
x x
x
0
0 0 cosθ
For a collision to occur, it is necessary that y y p T = at this time, or
( v
)⎛
0
0 0
v θ 0 0
2 2
2 2
sin
cos
θ
⎝ ⎜
⎞
⎠ ⎟− = − x g
t h
g
t
which reduces to
tanθ0
= h
x
0
Target
This requirement is satisfi ed provided that the gun is aimed at the initial location of the target.
Thus, a collision is guaranteed if the shooter aims the gun in this manner.
h
x0
q0

108 Chapter 3
3.60 (a) The components of the vectors are as follows:
Vector x-component (cm) y-component (cm)
d
1m 0 104
d
2m 46.0 19.5
d
1f 0 84.0
d
2f 38.0 20.2
The sums

d d d m 1m 2m = + and

d d d f 1f 2f = + are computed as
dm = (0 + 46 0) + (104 + 19 5) = 132 cm and = − 2 2 . . θ tan 1 104 19 5
0 460
69 6
+
+
⎛⎝ ⎜
⎞⎠ ⎟
.
= .
° .
df = (0 + 38 0) + (84 0 + 20 2) = 111 cm and = − 2 2 . . . θ tan 1 84 0 20 2
0 380
70 0
. .
.
.
+
+
⎛⎝ ⎜
⎞⎠ ⎟
= °
or

d d m f = 132 cm at 69.6° and = 111 cm at 70.0° .
(b) To normalize, multiply each component in the above calculation by the appropriate scale
factor. The scale factor required for the components of

d d 1m 2m and is
sm
= 200
cm
= 180 cm
1.11
and the scale factor needed for components of

d d 1f 2f and is
sf
= 200
cm
= 168 cm
1.19
After using these scale factors and recomputing the vector sums, the results are

d ′ = 146 cm at 69.6° and d ′ =
132 cm at 70.0°
m f
d d d m f.
The difference in the normalized vector sums is Δ ′ = ′ − ′
Vector x-component (cm) y-component (cm)
′
d
m 50.9 137
− ′
d
f –45.1 –124
Δ ′
d
Σx =5.74 Σy =12.8
Therefore, Δd′ = (Σx) + (Σy) = ( ) + ( ) = 2 2 2 2 5.74 12.8 cm 14.0 cm, and
θ =
⎛
1 1 . 12 8
⎝ ⎜
⎞
⎠ ⎟
= ⎛⎝
⎞⎠
.
.
tan− tan− = °
5 74
65 8
Σ
Σ
y
x
d
or Δ ′= 14 . 0 cm at 65.8
°

3.61 To achieve maximum range, the projectile should be
launched at 45° above the horizontal. In this case,
the initial velocity components are
0
2 x y = =
v v
v
0 0
The time of fl ight may be found from vy = v y − gt 0 by recognizing that when the projectile returns
to the original level, v v y y = − 0 .
Thus, the time of fl ight is
t
− −
v0 v0 v0 v0 v0 2 2
y y y =
g g g g
−
= = ⎛⎝ ⎜
⎞⎠ ⎟
=
2
2
The maximum horizontal range is then
v v v
2
= v ==
x g ⎛⎝ ⎜
g R t
⎞⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
0
0 0 0
2
2
[1]
Now, consider throwing the projectile straight upward at speed v0. At maximum height, vy = 0,
and the time required to reach this height is found from v v y y = − gt 0 as 0 0 = v − gt, which
yields t = v g 0 .
Therefore, the maximum height the projectile will reach is
2
( Δy ) = ( ) t
=
= max y av
g g + ⎛⎝
⎞⎠
⎛
⎝ ⎜
⎞
⎠ ⎟
v
0 v v v
2 2
0 0 0
Comparing this result with the maximum range found in equation [1] above reveals that
(Δy) R max = 2 provided the projectile is given the same initial speed in the two tosses.
If the girl takes a step when she makes the horizontal throw, she can likely give a higher initial
speed for that throw than for the vertical throw.
3.62 (a) x tx = v0 , so the time may be written as t x x = v0 .
Thus, y = v t − 1
gt 0
y 2
2 becomes
y
x
g
x
y
x x
=
⎛
⎝ ⎜
⎞
⎠ ⎟
−
⎛
⎝ ⎜
⎞
⎠ ⎟v
v v 0
0 0
2
1
2
or
y
g
v
v
x x
x
y
x
⎛
= −
⎝ ⎜
⎞
⎠ ⎟+
⎛
⎝ ⎜
⎞
⎠ ⎟
+
2
0
2
0
2 0
0 v
(b) Note that this result is of the general form y = ax2 + bx + c with
a
g
, b
, and c = 0
x
y
x
⎛
= −
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
2
2 0
0
v
v
0 v
45°
R
v0

110 Chapter 3
3.63 In order to cross the river in minimum time, the velocity of the boat
v
relative to the water (
BW) must be perpendicular to the banks
(and hence perpendicular to the velocity v
WS of the water relative
to shore).
The velocity of the boat relative to the water is v v v BW BS WS = − ,
where v
BS is the velocity of the boat relative to shore. Note that this vector
equation can be rewritten as v v v BS BW WS = + . Since v
BW and v
WS are to be perpendicular in this
case, the vector diagram for this equation is a right triangle with v
BS as the hypotenuse.
Hence, velocity of the boat relative to shore must have magnitude
v v v BS BW WS = 2 + 2 = ( km h)2 + ( km h)2 = km h 12 5.0 13
and be directed at
θ =
⎛
1 1 12
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠
tan− tan−
5 0
.
v
v
BW
WS
km h
km h⎟ = ° 67
to the direction of the current in the river (which is the same as the line of the riverbank).
The minimum time to cross the river is
1 5 60 . ⎞⎠
width of river km
min
1 h BW v
t= = ⎛⎝
12 km h
= 7 5. min
During this time, the boat drifts downstream a distance of
d = t = ( )( )⎛⎝
⎞⎠
vWS
3
km h min
h
60 min
10 m
5 0 7 5
1
. .
1 km
m
⎛
⎝ ⎜
⎞
⎠ ⎟
= 6.3 ×102

3.64 For the ball thrown at 45.0°, the time of fl ight is found from
1 2
2
Δy = v yt + ayt 0
as 0
− v
=g
0
t
t
2 1 1
⎛⎝ ⎜
⎞⎠ ⎟
2 2
which has the single nonzero solution of
t
0 2 = v
g 1
The horizontal range of this ball is
v v v
2 = =⎛⎝ ⎜
v =
R t
2
0 0 0
g g 1 0 x 1
2
⎞⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟
Now consider the fi rst arc in the motion of the second ball, started at angle q with initial speed v.
0Applied to this arc, Δy = v t + 1
at 2 becomes
0
y 2
y 0
g
t
= (v sinθ )t − 2
2 0 21 21
with nonzero solution
t
0 2 = v sinθ
g 21
Similarly, the time of fl ight for the second arc (started at angle q with initial speed v0 2) of this
ball’s motion is found to be
t
0 0 2 2
= ( ) =
v sinθ v sinθ
g g 22
The horizontal displacement of the second ball during the fi rst arc of its motion is
= =( )⎛
v v
v v =
R t
2 2 2
0 0
⎝ ⎜
g 21 0 x 21 0
⎞
⎠ ⎟
cos
sin sin c
θ
θ ( θ osθ ) v2 sin ( 2
θ )
= 0
g g
Similarly, the horizontal displacement during the second arc of this motion is
R
( v 2 ) 2
sin ( 2 θ ) 2
sin( θ )
= 0
= 1 v 0
g g 22
4
2
The total horizontal distance traveled in the two arcs is then
= + = v sin( θ )
R R R
5
2 0
4
g 2 21 22
2
(a) Requiring that the two balls cover the same horizontal distance (that is, requiring that
R R 2 1 = ) gives
5
4
v 2
sin ( θ ) v2
2 0
0
=
g g
5 θ = , which yields 2θ = 53.1°, so θ = 26.6° is the required
This reduces to sin (2 ) 4
projection angle for the second ball.

112 Chapter 3
(b) The total time of fl ight for the second ball is
= + = 2 v0 sinθ + v0 sinθ = 3v0 sinθ
t t t
g g g 2 21 22
Therefore, the ratio of the times of fl ight for the two balls is
t
t
g
g
2
1
0
0
3
2
3
2
( v
sin
)
= ( ) =
v
sin
θ
θ
With q = 26.6° as found in (a), this becomes
t
t
2
1
3
2
= sin(26.6°) = 0.950
3.65 The initial velocity components for the daredevil are v v 0 0 45 x= cos ° and v v 0 0 45 y= sin °, or
v v
v
0 0
0
2
25 0
x y = = = . ms
2
The time required to travel 50.0 m horizontally is
t
= Δ x
=( 50 0
) =
v25 0
0
x
2 2
.
.
m 2
m s
s
The vertical displacement of the daredevil at this time, and the proper height above the level of
the cannon to place the net, is
v ( )− 0
Δy t at y y = + = ⎛⎝ ⎜
⎞⎠ ⎟
1 2
2
25 0
2
2 2
1
2
9 80
.
.
m s
s ( m s2 )(2 2 s) 2
= 10 .
8 m
3.66 The vertical component of the salmon’s velocity as it leaves the water is
v v 0 0 6 26 45 0 4 43 y = + sinθ = +( . m s)sin . ° = + . m s
When the salmon returns to water level at the end of the leap, the vertical component of velocity
will be v v y y = − = − 0 4.43 m s.
The time the salmon is out of the water is given by
t
y y
= 0 903
a
y
1
0 4 43 4 43
9 80
−
=− −
−
=
v v . .
.
.
m s m s
m s2 s
The horizontal distance traveled during the leap is
L t t x = v = (v ) = ( ) ° 0 1 0 1 cosθ 6.26 m s cos45.0 (0.903 s) = 4.00 m
To travel this same distance underwater, at speed v = 3.58 m s, requires a time of
t
L
2
4 .
00
= = = 1 .
12
v
m
3.58 m s
s
The average horizontal speed for the full porpoising maneuver is then
vav
Δ ( )
total
= Δ
total
m
2 24 00
s
= =
+
x
t
L
0 903 1 2
t t
.
. +
=
1 12
3 96
.
.
s
m s

3.67 (a) and (b)
Since the shot leaves the gun horizontally, v0 x v0 = and the time required to reach the
target is
t
= Δ x =
x
v v 0 x
0
.
The vertical displacement occurring in this time is
Δy y t at g x
y y = − = + = −
⎛
⎝ ⎜
⎞
⎠ ⎟
v
2
v 0
0
2
1
2
0
1
2
which gives the drop as
= = 1
2 2 0
g =
y g
x
with , where v is the muzzle velocity
Ax A
⎛
⎝ ⎜
⎞
⎠ ⎟
2
2
2 0 v v
0
(c) If x = 3.00 m, and y = 0.210 m, then
0 210 m
2
A m 1
3.00 m
y
x
= =( ) = 2 33 × 10
− −
2 2
.
.
and
9 80
( × ) = − −
= = 14 5
v0 2 2
2 2 33 10
g
A
.
.
.
m s
m
m s
2
1
3.68 The velocity of the wind relative to the boat, v
WB, is given by v v v WB WE BE = − , where v
WE and
v
BE are the velocities of the wind and the boat relative to Earth, respectively. Choosing the
positive x-direction as east and positive y as north, these relative velocities have components of
v
WE ( ) = + knots x 17
v
WE ( ) = knots y 0
v
BE ( ) = x 0
v
BE ( ) = + knots y 20
so
v v v WB WE BE ( ) = ( ) − ( ) = + knots x x x 17
v v v WB WE BE ( ) = ( ) − ( ) = − knots y y y 20
Thus,
v v v WB WB WB = ( ) + ( ) = ( knots) + − knot x y
2 2 2 17 ( 20 s) = knots 2 26
and
θ =
( )
( )
⎛
⎝ ⎜⎞
tan− = − −
1 WB knots
⎠ ⎟
WB

v
v
y
x
tan 1 20
17
50
knots
⎛⎝
⎞⎠
= − °
or
v
WB = 26 knots at 50° south of east
The component of this velocity parallel to the motion of the boat (that is, parallel to a north-south
line) is ( ) . v
WB knots, or knots south y = −20 20

114 Chapter 3
3.69 (a) Take the origin at the point where the ball is launched. Then x0 = y0 = 0, and the
coordinates of the ball at time t later are
1 2
2 2
x t t x = v = (v ) 0 0 0 cosθ and y t at t g
t y y = + =( ) −⎛⎝
⎞⎠
2
v v 0
sinθ
0 0
When the ball lands at x = R = 240 m, the y-coordinate is y = 0 and the elapsed time is
found from
0
2 = ( ) −⎛⎝
2 0 0
⎞⎠
v sinθ t
g
t
for which the nonzero solution is
t
= 20 0 v sinθ
g
Substituting this time into the equation for the x-coordinate gives
x
2
sin θ
g g
= =( )⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
240
2
0 0
0 0 0
m v
v v
cos
θ
2
⎛
v ( )
g
( 2 sinθ cosθ ) = 0
sin 2 θ
0 0
0 ⎝ ⎜
⎞
⎠ ⎟
Thus, if v0 = 50.0 m s, we must have
sin
m g
( m)( .
m s2 ) v ( .
m s) = + 0.941
2 ( ) ( ) 2
θ = =
2 240 240 9 80
50 0 0
0
with solutions of 2 70 2 2 180 70 2 109 8 0 0 θ = . ° and θ = ° − . ° = . °.
So, the two possible launch angles are θ θ 0 0 = 35.1° and = 54.9° .
(b) At maximum height, vy = 0 and the elapsed time is given by
t
( ) −
v y v0 y
v
0 0 0 sinθ
a g
y
peak
= peak =
or
−
−
peak t
= v0 0 sinθ
g
The y coordinate of the ball at this time will be
sinθ g
2 sin 2
θ sin θ
g
θ θ peak peak
y t
g
t max sin sin = ( ) −⎛⎝
⎞⎠
v = (v )
v
0 0
2
0 0 2
0 0 0
0
2
2 2
0
0
2 g
2
⎛
⎝ ⎜
⎞
⎠ ⎟
−⎛⎝
⎞⎠
= v v
g
The maximum heights corresponding to the two possible launch angles are
( )
( 50 . 0 m s
) 2 sin 2 ( 35 .
1
°)
= 42 ( ) =
. max y 1
2 9 80
m s2 .2 m
and
( )
( 50 . 0 m s
) 2 sin 2 ( 54 .
9
°)
= 85 ( ) =
. max y 2
2 9 80
m s2 .4 m

3.70 (a) Consider the falling water to consist of droplets, each following a projectile trajectory. If
the origin is chosen at the level of the pool and directly beneath the end of the channel, the
parameters for these projectiles are
x0 = 0 y h 0 = = 2.35 m
v0 0 75 x = . ms v0 0 y =
ax = 0 a g y = −
The elapsed time when the droplet reaches the pool is found from y y t at y y − = + 0 0
v 2 as
1
2
g
t p or t
− h = − 2
0 0
2
h
g p = 2
The distance from the wall where the water lands is then
= = = h
=( .
) ( )
max x p x g R x t
.
.
v v 0 0
2
0 750
2 2 35
9 80
m s
m
m s
m 2 = 0.519
This space is too narrow for a pedestrian walkway.
(b) It is desired to build a model in which linear dimensions, such as the height hmodel and
horizontal range of the water Rmodel, are one-twelfth the corresponding dimensions in
the actual waterfall. If vmodel is to be the speed of the water fl ow in the model, then we
would have
= v ( ) = v model
h
g model model p model model
R t
2
or
vmodel model
g
h
model
R g
actual
actual
= = ( R
2 12 2 h 12) =
⎛
= 1
12 2 12
⎝ ⎜
⎞
⎠ ⎟
g
h actual
R
actual
actual v
and the needed speed of fl ow in the model is
v
v
model
actual m s
= = = m s
12
0 750
12
0 217
.
.

116 Chapter 3
3.71 (a) Applying Δy = v 0
yt + 1
t 2
ay2 to the vertical motion of the fi rst snowball gives
. m s sin . ° t + (− . m s2 )t
2 = ( ) ⎡⎣
0 25 0 70 0
1
2
9 80 1 1
⎤⎦
which has the nonzero solution of
t1
2 ( 25 0 ) 70 0
°
= 4 79 9 80
=
. sin .
.
.
m s
m s
s 2
as the time of fl ight for this snowball.
The horizontal displacement this snowball achieves is
Δx tx = = ( ) ⎡⎣
v ° ( ) = 0 1 25.0 m s cos70.0 4.79 s 41.0 m
⎤⎦
Now consider the second snowball, also given an initial speed of v= 25.0 m s, thrown at
0 angle q , and is in the air for time t. Applying Δy = v t + 1
at 20
y 2
y 2 to its vertical motion yields
. m s sinθ t + (− . m s2 )t
2 = ( ) ⎡⎣
0 250
1
2
9 80 2 2
⎤⎦
which has a nonzero solution of
t2
2 ( 250
. ) sin
= = ( 5 10 ) 9 .
80
. sin
m s
m s
θ
s 2
θ
We require the horizontal range of this snowball be the same as that of the fi rst ball, namely
v [( ) ] = 0 2 25.0 m s cosθ 5.10 s sinθ 41.0 m
Δx tx = = ( ) ⎡⎣
⎤⎦
This yields the equation
sin
= 41 0
m
( = . 25 0 m s )( 5 10
s
) θ cosθ
.
. .
0 321
From the trigonometric identity sin 2θ = 2sinθ cosθ , this result becomes
sin 2θ = 2(0.321) = 0.642
so
2θ = 40.0°
and the required angle of projection for the second snowball is
θ = 20.0° above the horizontal
(b) From above, the time of fl ight for the fi rst snowball is t1 = 4.79 s and that for the second
snowball is
t2 = (5.10 s)sinθ = (5.10 s)sin 20.0° = 1.74 s
Thus, if they are to arrive simultaneously, the time delay between the fi rst and second
snowballs should be
Δt = t − t = − = 1 2 4.79 s 1.74 s 3.05 s

3.72 First, we determine the velocity with which the dart leaves the gun by using the data collected
when the dart was fi red horizontally (v= 0) from a stationary gun. In this case, Δy = v + 1
t 0 y 0
yt 2
ay2
gives the time of fl ight as
t
y
ay
= =
(− )
−
= 2 2 100
0 452
Δ .
.
m s
9.80 m s
s 2
Thus, the initial speed of the dart relative to the gun is
Δ .
5 00
= = = = m s 0
v v DG
m
11 1 x
0.452 s
x
t
.
At the instant when the dart is fi red horizontally from a moving
gun, the velocity of the dart
relative to the gun may be expressed
as v v v DG DE GE = − where v v DE GE and are the velocities of the
dart and gun relative to Earth respectively. The initial velocity
of the dart relative to Earth is therefore
v v v v 0 DE DG GE = = +
From the vector diagram, observe that
v v 0 45 0 2 00 45 0 1 41 y = − °= −( ) ° = − GE sin . . m s sin . . m s
and
v v v 0 45 0 11 1 2 00 45 x= + °= +( ) DG GE cos . . m s . m s cos .0° = 12.5 m s
The vertical velocity of the dart after dropping 1.00 m to the ground is
v v y y y = − + a y = − (− ) + (− ) − 0
2 2 2 Δ 1.41 m s 2 9.80 m s2 ( 1.00 m) = −4.65 m s
and the time of fl ight is
t
y y
a
y
=
−
=
− − (− )
−
=
v v0 4 65 1 41
9 80
0 3
. .
.
.
m s m s
m s2 30 s
The displacement during the fl ight is Δx tx = v = ( )( ) = 0 12.5 m s 0.330 s 4.12 m .

118 Chapter 3
3.73 (a) First, use Δ x = v xt + 1
axt 0
2
2 to fi nd the time for the coyote to travel 70 m, starting from rest
with constant acceleration ax = 15 m s2:
t
= = 3 1 ( ) Δ m =
x
ax
1
2 270
15
m s
s 2 .
The minimum constant speed the roadrunner must have to reach the edge in this time is
v = = = Δx
t1
70
23
m
3.1 s
m s
(b) The initial velocity of the coyote as it goes over the edge of the cliff is horizontal and
equal to
v v 0 0 1 = = 0 + = (15 )(3 1 ) = 46 x x a t m s2 . s m s
From Δy = v t + 1
at 0
y 2
y 2, the time for the coyote to drop 100 m with v0 0 y = is
t
(− )
= Δ = m =
4 52 y
ay
2
2 2 100
−
s . 2
9 80
m s
.
The horizontal displacement of the coyote during his fall is
Δx = v t + 1
at 2 = ( )( ) + 1
( ) 0 x 2
2 x 2
2 46 m s 4.52 s 15 m s2 4 52 3 6 10 ( . s)2 = . × 2 m

Solucionario Fundamentos de Física 9na edición Capitulo 3

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Solucionario Fundamentos de Física 9na edición Capitulo 3