This document outlines the agenda and content for Physics 111 Lecture 2. The lecture will recap 1-D motion with constant acceleration, cover 1-D free fall, review vectors, and discuss 3-D kinematics. It provides examples and problems involving calculating the time and velocity of objects in free fall. It also demonstrates how to use vectors to represent position, velocity, and acceleration in 3 dimensions and how to add vectors. The components of motion in different directions are treated independently for 3-D problems.

1. Physics 111: Lecture 2, Pg 1
Physics 111: Lecture 2
Today’s Agenda
 Recap of 1-D motion with constant acceleration
 1-D free fall
example
 Review of Vectors
 3-D Kinematics
Shoot the monkey
Baseball
Independence of x and y components

2. Physics 111: Lecture 2, Pg 2
Review:
 For constant acceleration we found:
atvv 0 
2
00 at
2
1
tvxx 
consta 
x
a
v
t
t
t
v)(v
2
1
v
)x2a(xvv
0av
0
2
0
2


 From which we derived:

3. Physics 111: Lecture 2, Pg 3
Recall what you saw:
2
00 at
2
1
tvxx 

4. Physics 111: Lecture 2, Pg 4
1-D Free-Fall
 This is a nice example of constant acceleration (gravity):
 In this case, acceleration is caused by the force of gravity:
Usually pick y-axis “upward”
Acceleration of gravity is “down”:
y
ay =  g
gtvv
y
0y 
2
y00 tg
2
1
tvyy 
y
a
v
t
t
t
gay 

5. Physics 111: Lecture 2, Pg 5
Gravity facts:
 g does not depend on the nature of the material!
Galileo (1564-1642) figured this out without fancy clocks
& rulers!
 demo - feather & penny in vacuum
 Nominally, g = 9.81 m/s2
At the equator g = 9.78 m/s2
At the North pole g = 9.83 m/s2
 More on gravity in a few lectures!
Penny
& feather
Ball
w/ cup

6. Physics 111: Lecture 2, Pg 6
Problem:
 The pilot of a hovering helicopter
drops a lead brick from a height
of 1000 m. How long does it take
to reach the ground and how fast
is it moving when it gets there?
(neglect air resistance)
1000 m

7. Physics 111: Lecture 2, Pg 7
Problem:
 First choose coordinate system.
Origin and y-direction.
 Next write down position equation:
 Realize that v0y = 0.
2
0y0 gt
2
1
tvyy 
2
0 gt
2
1
yy 
1000 m
y = 0
y

8. Physics 111: Lecture 2, Pg 8
Problem:
 Solve for time t when y = 0 given
that y0 = 1000 m.
 Recall:
 Solve for vy:
y0 = 1000 m
y
s314
sm819
m10002
g
y2
t 2
0 .
.



2
0 gt
2
1
yy -
y = 0
)( 0
2
y0
2
y yya2vv -- 
sm140
gy2v 0y
/


9. Physics 111: Lecture 2, Pg 9
Lecture 2, Act 1
1D free fall
 Alice and Bill are standing at the top of a cliff of height
H. Both throw a ball with initial speed v0, Alice straight
down and Bill straight up. The speed of the balls when
they hit the ground are vA and vB respectively. Which
of the following is true:
(a) vA < vB (b) vA = vB (c) vA > vB
v0
v0
BillAlice
H
vA vB

10. Physics 111: Lecture 2, Pg 10
Lecture 2, Act 1
1D Free fall
 Since the motion up and back down is symmetric, intuition
should tell you that v = v0
We can prove that your intuition is correct:
v0
Bill
H
v = v0
  0HHg2vv 2
0
2
 )(Equation:
This looks just like Bill threw
the ball down with speed v0, so
the speed at the bottom should
be the same as Alice’s ball.
y = 0

11. Physics 111: Lecture 2, Pg 11
Lecture 2, Act 1
1D Free fall
 We can also just use the equation directly:
 H0g2vv 2
0
2
 )(Alice:
v0
v0
Alice Bill
y = 0
 H0g2vv 2
0
2
 )(Bill:
same !!

12. Physics 111: Lecture 2, Pg 12
Vectors (review):
 In 1 dimension, we could specify direction with a + or - sign.
For example, in the previous problem ay = -g etc.
 In 2 or 3 dimensions, we need more than a sign to specify the
direction of something:
 To illustrate this, consider the position vector r in 2 dimensions.
Example: Where is Chicago?
 Choose origin at Urbana
 Choose coordinates of
distance (miles), and
direction (N,S,E,W)
 In this case r is a vector that
points 120 miles north.
Chicago
Urbana
r

13. Physics 111: Lecture 2, Pg 13
Vectors...
 There are two common ways of indicating that something is
a vector quantity:
Boldface notation: A
“Arrow” notation:
A =

A

A

14. Physics 111: Lecture 2, Pg 14
Vectors...
 The components of r are its (x,y,z) coordinates
 r = (rx ,ry ,rz ) = (x,y,z)
 Consider this in 2-D (since it’s easier to draw):
rx = x = r cos 
ry = y = r sin 
y
x
(x,y)

where r = |r |
r arctan( y / x )

15. Physics 111: Lecture 2, Pg 15
Vectors...
 The magnitude (length) of r is found using the Pythagorean
theorem:
r   r x y2 2
r
y
x
 The length of a vector clearly does not depend on its direction.

16. Physics 111: Lecture 2, Pg 16
Unit Vectors:
 A Unit Vector is a vector having length 1
and no units
 It is used to specify a direction
 Unit vector u points in the direction of U
Often denoted with a “hat”: u = û
 Useful examples are the Cartesian
unit vectors [ i, j, k ]
 point in the direction of the
x, y and z axes
U
x
y
z
i
j
k
û

17. Physics 111: Lecture 2, Pg 17
Vector addition:
 Consider the vectors A and B. Find A + B.
A
B
A B A B
C = A + B
 We can arrange the vectors as we want, as long as we
maintain their length and direction!!

18. Physics 111: Lecture 2, Pg 18
Vector addition using components:
 Consider C = A + B.
(a) C = (Ax i + Ay j) + (Bx i + By j) = (Ax + Bx)i + (Ay + By)j
(b) C = (Cx i + Cy j)
 Comparing components of (a) and (b):
 Cx = Ax + Bx
 Cy = Ay + By
C
BxA
By
B
Ax
Ay

19. Physics 111: Lecture 2, Pg 19
Lecture 2, Act 2
Vectors
 Vector A = {0,2,1}
 Vector B = {3,0,2}
 Vector C = {1,-4,2}
What is the resultant vector, D, from
adding A+B+C?
(a) {3,5,-1} (b) {4,-2,5} (c) {5,-2,4}

20. Physics 111: Lecture 2, Pg 20
Lecture 2, Act 2
Solution
D = (AXi + AYj + AZk) + (BXi + BYj + BZk) + (CXi + CYj + CZk)
= (AX + BX + CX)i + (AY + BY+ CY)j + (AZ + BZ + CZ)k
= (0 + 3 + 1)i + (2 + 0 - 4)j + (1 + 2 + 2)k
= {4,-2,5}

21. Physics 111: Lecture 2, Pg 21
3-D Kinematics
 The position, velocity, and acceleration of a particle in 3
dimensions can be expressed as:
r = x i + y j + z k
v = vx i + vy j + vz k (i , j , k unit vectors )
a = ax i + ay j + az k
 We have already seen the 1-D kinematics equations:
a
dv
dt
d x
dt
 
2
2
v
dx
dt
x x(t )

22. Physics 111: Lecture 2, Pg 22
3-D Kinematics
 For 3-D, we simply apply the 1-D equations to each of the
component equations.
 Which can be combined into the vector equations:
r = r(t) v = dr / dt a = d2r / dt2
a
d x
dt
x 
2
2
v
dx
dt
x 
x x(t )
a
d y
dt
y 
2
2
v
dy
dt
y 
y y t ( )
a
d z
dt
z 
2
2
v
dz
dt
z 
z z t ( )

23. Physics 111: Lecture 2, Pg 23
3-D Kinematics
 So for constant acceleration we can integrate to get:
a = const
v = v0 + a t
r = r0 + v0 t + 1/2 a t2
(where a, v, v0, r, r0, are all vectors)

24. Physics 111: Lecture 2, Pg 24
2-D Kinematics
 Most 3-D problems can be reduced to 2-D problems when
acceleration is constant:
Choose y axis to be along direction of acceleration
Choose x axis to be along the “other” direction of
motion
 Example: Throwing a baseball (neglecting air resistance)
Acceleration is constant (gravity)
Choose y axis up: ay = -g
Choose x axis along the ground in the direction of the
throw
lost marbles

25. Physics 111: Lecture 2, Pg 25
“x” and “y” components of motion are
independent.
 A man on a train tosses a ball straight up in the air.
View this from two reference frames:
Reference frame
on the moving train.
Reference frame
on the ground.
Cart

26. Physics 111: Lecture 2, Pg 26
Problem:
 Mark McGwire clobbers a fastball toward center-field. The
ball is hit 1 m (yo ) above the plate, and its initial velocity is
36.5 m/s (v ) at an angle of 30o () above horizontal. The
center-field wall is 113 m (D) from the plate and is 3 m (h)
high.
What time does the ball reach the fence?
Does Mark get a home run?

v
h
D
y0

27. Physics 111: Lecture 2, Pg 27
Problem...
 Choose y axis up.
 Choose x axis along the ground in the direction of the hit.
 Choose the origin (0,0) to be at the plate.
 Say that the ball is hit at t = 0, x = x0 = 0
Equations of motion are:
vx = v0x vy = v0y - gt
x = vxt y = y0 + v0y t - 1/ 2 gt2

28. Physics 111: Lecture 2, Pg 28
Problem...
 Use geometry to figure out v0x and v0y :
y
x
g

v
v0x
v0y
Find v0x = |v| cos .
and v0y = |v| sin .
y0

29. Physics 111: Lecture 2, Pg 29
Problem...
 The time to reach the wall is: t = D / vx (easy!)
 We have an equation that tell us y(t) = y0 + v0y t + a t2/ 2
 So, we’re done....now we just plug in the numbers:
 Find:
 vx = 36.5 cos(30) m/s = 31.6 m/s
 vy = 36.5 sin(30) m/s = 18.25 m/s
 t = (113 m) / (31.6 m/s) = 3.58 s
 y(t) = (1.0 m) + (18.25 m/s)(3.58 s) -
(0.5)(9.8 m/s2)(3.58 s)2
= (1.0 + 65.3 - 62.8) m = 3.5 m
 Since the wall is 3 m high, Mark gets the homer!!

30. Physics 111: Lecture 2, Pg 30
Lecture 2, Act 3
Motion in 2D
 Two footballs are thrown from the same point on a flat field.
Both are thrown at an angle of 30o above the horizontal.
Ball 2 has twice the initial speed of ball 1. If ball 1 is caught
a distance D1 from the thrower, how far away from the
thrower D2 will the receiver of ball 2 be when he catches it?
(a) D2 = 2D1 (b) D2 = 4D1 (c) D2 = 8D1

31. Physics 111: Lecture 2, Pg 31
Lecture 2, Act 3
Solution
 The distance a ball will go is simply
x = (horizontal speed) x (time in air) = v0x t
2
y00 tg
2
1
tvyy 
 To figure out “time in air”, consider the
equation for the height of the ball:
0tg
2
1
tv 2
y0  When the ball is caught, y = y0
0tg
2
1
vt y0 





 g
v
2t
y0

t  0 (time of throw)
(time of catch)
two
solutions

32. Physics 111: Lecture 2, Pg 32
Lecture 2, Act 3
Solution
g
v
2t
y0
 So the time spent in the air is proportional to v0y :
 Since the angles are the same, both v0y and v0x for ball 2
are twice those of ball 1.
ball 1
ball 2
v0y ,1
v0x ,1
v0y ,2
v0x ,2
v0,1
v0,2
 Ball 2 is in the air twice as long as ball 1, but it also has twice
the horizontal speed, so it will go 4 times as far!!
x = v0x t

33. Physics 111: Lecture 2, Pg 33
Shooting the Monkey
(tranquilizer gun)
 Where does the zookeeper
aim if he wants to hit the monkey?
( He knows the monkey will
let go as soon as he shoots ! )

34. Physics 111: Lecture 2, Pg 34
Shooting the Monkey...
 If there were no gravity, simply aim
at the monkey
r = r0
r =v0t

35. Physics 111: Lecture 2, Pg 35
Shooting the Monkey...
r = v0 t - 1/2 g t2
 With gravity, still aim at the monkey!
r = r0 - 1/2 g t2
Dart hits the
monkey!

36. Physics 111: Lecture 2, Pg 36
Recap:
Shooting the monkey...
x = x0
y = -1/2 g t2
 This may be easier to think about.
It’s exactly the same idea!!
x = v0 t
y = -1/2 g t2

37. Physics 111: Lecture 2, Pg 37
Recap of Lecture 2
 Recap of 1-D motion with constant acceleration. (Text: 2-3)
 1-D Free-Fall (Text: 2-3)
example
 Review of Vectors (Text: 3-1 & 3-2)
 3-D Kinematics (Text: 3-3 & 3-4)
Shoot the monkey (Ex. 3-11)
Baseball problem
Independence of x and y components
 Look at textbook problems Chapter 3: # 35, 39, 69, 97