kinematics in 2 dimensions

1. Newtonian Mechanics
Kinematics
in Two Dimensions

2. Projectile
is any object that once
projected or dropped
continues in motion by its
own inertia and is influenced
only by the downward force of
gravity.

5. is an
idealized kind of motion that
occurs when a moving object
(the projectile) experiences
only the acceleration due to
gravity, which acts vertically
downward.
Projectile motion

6. Projectile motion
If the trajectory of the projectile is
near the earth’s surface, the vertical
component ay of the acceleration has
a magnitude of 9.80 m/s2. The
acceleration has no horizontal
component (aₓ " 0 m/s2), the effects of
air resistance being negligible.

7. Gravity acts to influence the
vertical motion of projectile, this
causing a vertical acceleration.
The Horizontal motion of the
projectile is the result of the
tendency of any object in motion
to remain in motion at Constant
Velocity.

8. Inertia
 Tendency of an object to resist
 Also affects the projectile
motion.
 If there is no air resistance and
no gravity, the object has a
constant velocity because of
INERTIA.

9. Types of Projectile
 Falling
 Toss upward
 Toss upward with an angle

12. Ahhh the memories…

13. o
d

 initialposition
d

 finalposition
o
d

 d

d

 displacement
Displacement

14. o

t to t

d
v 
  
 d  d
Average velocity is the
displacement divided by
the elapsed time.
Velocity

15. t to t

v

a


v

 v

o
o
v

v
 v

Acceleration
DEFINITION OF AVERAGE ACCELERATION

16. 2
x  x  v t  1 at2
v  vo  at
o
o
o o
v  v  2ax  x 
2
2
Quick Review
Equations of Kinematics

17. Keys to Solving 2-D Problems
1. Resolve all vectors into components


x-component
Y-component
2. Work the problem as two one-dimensional
problems.
 Each dimension can obey different
equations of motion.
3. Re-combine the results for the two components
at the end of the problem.

18. x
ox
x
v  v  a t
v2
 v2
 2a x  x 
x ox x o
2 x
t  1 a t2
x  xo  vo, x
x component

19. 2 y
o oy
y  y  v t  1 a t 2
v2
 v2
 2a y  y 
y oy y o
y component
vy  voy  ayt

20. Actual motion
 The x part of the motion occurs exactly as it would if the y
part did not occur at all, and vice versa.
 Remember the bullet drop video!

21. In the x direction, the spacecraft has an initial velocity
component of +22 m/s and an acceleration of +24 m/s2. In
the y direction, the analogous quantities are +14 m/s and
an acceleration of +12 m/s2. Find (a) x and vx, (b) y and
vy, and (c) the final velocity of the spacecraft at time 7.0 s.
A Moving Spacecraft
Ex a m ple 1

22. In the x direction, the spacecraft has an initial velocity
component of +22 m/s and an acceleration of +24 m/s2. In the
y direction, the analogous quantities are +14 m/s and an
acceleration of +12 m/s2. Find (a)  x and vx, (b)  y and vy,
and (c) the final velocity of the spacecraft at time 7.0 s.
x ax vx vox t
? +24.0 m/s2 ? +22 m/s 7.0 s
y ay vy voy t
? +12.0 m/s2 ? +14 m/s 7.0 s
Ex a mple 1 A Moving Spacecraft

23. x ax vx vox t
? +24.0 m/s2 ? +22 m/s 7.0 s
740 m
2
2
1
2
2
24m s 7.0s 
 22m s7.0s
x  v t  1 a t2
ox x
vx  vox axt
vx  22m s 24ms2
7.0s 190m s
x component

24. y ay vy voy t
? +12.0 m/s2 ? +14 m/s 7.0 s
390 m
2
2
2
1
2
 12m s 7.0s 
 14m s7.0s
y  v t  1
a t2
oy y
vy  voy  ayt
 14m s 12m s2
7.0s 98m s
y component

25.  210 m s
vx  1 9 0 m s
v  190 m s2
 98 m s2
  tan1
98190  27
Resultant velocity
v
v y  98 m s


26. Projectile Motion





Something is fired, thrown, shot, or hurled near the
Earth’s surface
Horizontal velocity is constant
•ax = 0, vo,x = vx
Vertical velocity is accelerated
•ay = g = 9.80665 m/s2 ≈ 10 m/s2 (constant)
Air resistance is ignored
If launch angle is not horizontal or vertical
•must resolve initial velocity into x and y components

27. The airplane is moving horizontally with a constant velocity of
+115 m/s at an altitude of 1050 m. Determine the time required
for the care package to hit the ground.
Ex a mple 3 A Falling Care Package

28. y ay vy voy t
-1050 m -9.80 m/s2 0 m/s ?

29. y ay vy voy t
-1050 m -9.80 m/s2 0 m/s ?
2
oy y
y  v t  1 a t 2
2 y
Δy  1 a t 2
ay
t 
2y
0

30. E x a m p l e 4
The Velocity of the Care Package
A plane is moving horizontally with a constant
velocity of 115 m/s at an altitude of 1050 m.
Ignoring air resistance, determine the velocity of
a package just before it hits the ground.

31. v y
y
2
o,y
2
 2a Δy
v 
v  v2
v2
x y
v   v2
 2ay
v2
o,x o,y
vx vo, x
v  115m s2
 0m s2
 29.81m s2
1050m
v  184 ms
v
cos 
vx
 
1  vx 
  cos
  51.3o
y ay v vo,y vo,x t
-1050 m -9.80 m/s2 ? 0 m/s 115 m/s 14.6 s


   
 

v 184 m s
 115m s 

32. Conceptual Example 5
I Shot a Bullet into the Air…
Suppose you are driving a
convertible with the top
down. The car is moving to
the right at constant velocity.
You point a rifle straight up
into the air and fire it. In the
absence of air resistance,
where would the bullet land –
(a) behind you
(b) ahead of you
(c) in the barrel of the rifle

33. A placekicker kicks a football at and angle of 40.0 degrees and
the initial speed of the ball is 22 m/s. Ignoring air resistance,
determine the maximum height that the ball attains.
Ex a m p le 6 The Height of
a Kickoff

34. voy
vo

vox
voy  vo sin  22m ssin40
 14m s
vox  vo sin  22m scos40
 17m s

35. y ay vy voy t
? -9.80 m/s2 0 14 m/s

36. 3.3 Project ile Motion
y
 2a Δy
v2
 v2
y o,y
y ay vy voy t
? -9.80 m/s2 0 14 m/s
v2
v2
y oy
2ay
y 
0  14m s2
y 
29.8m s2
  10m

37. What is the time of flight between kickoff and landing?
Ex a m p le 7
The Time of Flight of a
Kickoff

38. y ay vy voy t
0 -9.80 m/s2 14 m/s ?
2 y
o,yt  1 a t2
y  v
t  2.9 s
y
1
2
o,y t  0s
y  0  tv  a t 
a t
y
1
2
0  vo,y 
y
a 9.80m s2
t  
2vo,y
 
214m s

39. Calculate the range R of the projectile.
Ex a m p le 8
The Range of a Kickoff

40. x ax vx Vo,x t
? 0 17 m/s 2.9 s
2 x
o,xt  1 a t2
x v
x  17m s2.9s
x  49m
0

41. Question #9
A diver running 1.8 m/s dives out horizontally from
the edge of a vertical cliff and 3.0 s later reaches the
water below. How high was the cliff?
h
d
v

42. Question #10
A diver running 1.8 m/s dives out horizontally from the
edge of a vertical cliff and 3.0 s later reaches the water
below. How far from its base did the diver hit the water?
h
d
v

43. Question #11
When a football in a field goal attempt reaches
its maximum height, how does its speed
compare to its initial speed?
a) It is zero.
b) It is equal to its initial speed.
c) It is greater than its initial speed.
d) It is less than its initial speed.

44. Question #12
The Zambezi River flows over Victoria Falls in Africa. The
falls are approximately 108 m high. If the river is flowing
horizontally at 3.6 m/s just before going over the falls,
what is the speed of the water when it hits the bottom?
Assume the water is in freefall as it drops.

45. Question #13
An astronaut on the planet Zircon tosses a rock horizontally
with a speed of 6.75 m/s. The rock falls a distance of 1.20 m
and lands a horizontal distance of 8.95 m from the astronaut.
What is the acceleration due to gravity on Zircon?

46. For The Adventurous
 Here are some more advanced equations.
 They are not given on the equation sheet

 You are free to use them if they are “memorized”
•You can have programs stored in your calc
Unlikely you will need them, but…

47. Vertical Position Equation

vo h
x
y
Vertical Position as a function
of vo, , and horizontal position
vo, y  vo sin
2
1
y  vo, yt 
2
gt
2
1
o,x x
x  v t  a t2
vo, x  vo cos
x
vo cos
t 
2
1














 o
 o
o
v cos
x
 g
2
v cos
x
y  v sin
2 2
o
2v cos 
g x2
y  xtan 
0

48. Maximum Height
ttop
v sin
 o
g

vo h
Equation for the maximum
Height of a projectile.
At max height,
vy = 0
vy  vo sin   gt  0
2
2
1
gt
o
y  v sin t


  g
g 2
g 
o

 v sin  1  v sin 
2
y  vo sin
o

2g
v2
sin2

ymax  o

49. x  vo cost 2
1
2
gt
o
y  v sin t
2v sin
t o
g




 g
o
 2vo sin
x  v cos
sin2u 2sinu cosu

2
g
v
x  o
sin 2
sin cos
2
2vo
g
x 
 0

Range Equation vo
d

50. Chapter 3
Motion in Two Dimensions
Section 4:
Relative Velocity

51. Relative Motion Problems
 Describe motion of an object from some fixed point
 Relative motion problems are difficult to do unless one applies
vector addition concepts
 Define a vector for an object velocity relative to the fluid, and
another for the velocity of the fluid relative to the observer.
 Add two vectors together.

52. v

 v

 v

PG PT TG
velocity of person velocity of person
relative to the ground relative to the train
velocity of train
relative to the ground

53. The engine of a boat drives it across a river that is 1800m wide.
The velocity of the boat relative to the water is 4.0m/s directed
perpendicular to the current. The velocity of the water relative
to the shore is 2.0m/s.
(a)What is the velocity of the
boat relative to the shore?
(b)How long does it take for
the boat to cross the river?
Ex a m p le 11 Crossing a River

54. BS
v  4.0m s2
 2.0m s2
BS BW WS
v

 v

 v

 

2.0

x vBW vWS vBS t
1800m 4.0m/s 2.0m/s ? ?
BW WS
BS
v2
 v2
v 
 4.5 m s
vBS
 WS 

 v


1 vBW 
 tan
  tan1 4.0   63

55. 4.0 m s
x vBW vWS t
1800m 4.0m/s 2.0m/s ?
vBW
x
t 
t 
1800 m
 450 s
2
1
BW
x  v t 
0
at2

56. Question #14
You are trying to cross a river that flows due south
with a strong current. You start out in your motorboat
on the east bank desiring to reach the west bank
directly west from your starting point. You should
head your motorboat
a) due west.
b) due north.
c) in a southwesterly direction.
d) in a northwesterly direction.

57. Question #15
At an air show, three planes are flying horizontally due
east. The velocity of plane A relative to plane B is vAB;
the velocity of plane A relative to plane C is vAC; and the
velocity of plane B relative to plane C is vBC. Determine
vAB if vAC = +10 m/s and vBC = +20m/s?
b) +10 m/s
d) +20 m/s
a) 10 m/s
c) 20 m/s
e) zero m/s

58. Question #16
A train is traveling due east at a speed of 26.8 m/s relative to
the ground. A passenger is walking toward the front of the
train at a speed of 1.7 m/s relative to the train. Directly
overhead the train is a plane flying horizontally due west at a
speed of 257.0 m/s relative to the ground. What is the
horizontal component of the velocity of the airplane with
respect to the passenger on the train?
b) 285.5 m/s, due west
d) 231.9 m/s, due west
a) 258.7 m/s, due west
c) 226.8 m/s, due west
e) 257.0 m/s, due west

59. Question #17
Sailors are throwing a football on the deck of an aircraft carrier as it is sailing with a
constant velocity due east. Sailor A is standing on the west side of the flight deck
while sailor B is standing on the east side. Sailors on the deck of another aircraft
carrier that is stationary are watching the football as it is being tossed back and
forth as the first carrier passes. Assume that sailors A and B throw the football
with the same initial speed at the same launch angle with respect to the horizontal,
do the sailors on the stationary carrier see the football follow the same parabolic
trajectory as the ball goes east to west as it does when it goes west to east?
a)Yes, to the stationary sailors, the trajectory the ball follows is the same whether
it is traveling west to east or east to west.
b)No, to the stationary sailors, the length of the trajectory appears shorter as it
travels west to east than when it travels east to west.
c)No, to the stationary sailors, the ball appears to be in the air for a much longer
time when it is traveling west to east than when it travels east to west.
d)No, to the stationary sailors, the length of the trajectory appears longer as it
travels west to east than when it travels east to west.
e)No, to the stationary sailors, the ball appears to be in the air for a much shorter
time when it is traveling west to east than when it travels east to west.

60. Question #18
Cars A and B are moving away from each other as car A moves
due north at 25 m/s with respect to the ground and car B moves
due south at 15 m/s with respect to the ground. What are the
velocities of the other car according to the two drivers?
a)Car A is moving due north at 25 m/s; and car B is moving due
south at 15 m/s.
b)Car A is moving due south at 25 m/s; and car B is moving due
north at 15 m/s.
c)Car A is moving due north at 40 m/s; and car B is moving due
south at 40 m/s.
d)Car A is moving due south at 40 m/s and car B is moving due
north at 40 m/s.
e)Car A is moving due north at 15 m/s and car B is moving due
south at 25 m/s.

61. Question #19
John always paddles his canoe at constant speed v with respect
to the still water of a river. One day, the river current was due
west and was moving at a constant speed that was close to v
with respect to that of still water. John decided to see whether
making a round trip across the river and back, a north-south trip,
would be faster than making a round trip an equal distance east-
west. What was the result of John’s test?
a) The time for the north-south trip was greater than the time for
the east-west trip.
b) The time for the north-south trip was less than the time for the
east-west trip.
c) The time for the north-south trip was equal to the time for the
east-west trip.
d) One cannot tell because the exact speed of the river with
respect to still water is not given.

62. Question #20
A boat attempts to cross a river. The boat’s speed with
respect to the water is 12.0 m/s. The speed of the river
current with respect to the river bank is 6.0 m/s. At what
angle should the boat be directed so that it crosses the
river to a point directly across from its starting point?

 BW 

 v


1  vWS
  sin
6.0m/s

12m s

 
1 6.0m s 
  sin