User's Manual

®
Model No. ME-6830 Exp. 7: Vary the Angle to Maximize the Height
012-05375C 35
Exp. 7: Vary the Angle to Maximize the Height
Equipment Needed
Purpose
The purpose of this experiment is to find the launch angle that will maximize the height on a vertical wall for a
projectile launched at a fixed horizontal distance from the wall.
Theory
When the ball is shot at an angle at a fixed distance, x, from a
target such as a vertical wall, the ball hits the wall at a height y
given by:
where y
0
is the initial height of the ball, v
0
is the initial speed
of the ball as it leaves the muzzle,  is the angle of inclination
above horizontal, g is the acceleration due to gravity, and t is
the time of flight. The range is the horizontal distance, x,
between the muzzle of the Launcher and the place where the
ball hits, given by
Solving this equation for the time of flight, t, gives
Substituting for t in the equation for y gives
To find the angle, , that gives the maximum height, y, find the first derivative of the equation for y and set it equal
to zero. Solve for the angle, .
Solving for the angle, , gives:
Since the second derivative is negative for 
max
, the angle is a maximum. To find the initial speed of the ball, use
the fixed distance, x, and the maximum height, y
max
. Solve the y-equation for v
0
and plug in the values for y
max
,

max
, and x.
Item Item
Projectile Launcher and plastic ball Board to protect wall
Meter stick or measuring tape Sticky tape
White paper, large sheet Carbon paper (several sheets)
Plumb bob and string
θ
C
A
U
TIO
N!
D
O
N
O
T LO
O
K
D
O
W
N BAR
R
E
L!
C
A
U
TIO
N!
D
O
N
O
T LO
O
K
D
O
W
N BAR
R
E
L!
CAUTION!
DO NOT LOOK
DOWN THE
L
O
N
G
R
A
N
G
E
M
E
D
I
U
M
R
A
N
G
E
S
H
O
R
T
R
A
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G
E
P
o
s
i
t
i
o
n
o
f
B
a
l
l
L
a
u
n
c
h
SHOR
T RANGE
P
R
O
J
E
C
T
IL
E
L
A
U
N
C
H
E
R
ME-6800
Y
e
l
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w
B
a
n
d
i
n
W
i
n
d
o
w
In
d
ic
a
t
e
s
R
a
n
g
e
.
9
8
7
6
5
4
3
2
1
0
WEAR
SAFETY
GLASSES
WHEN IN USE.
Us
e 25 m
m
ba
lls
ON
L
Y
!
Figure 7.1: Maximizing Height
Range
Initial
height
Angle
Initial
speed
y
y
0
v
0
x
yy
0
v
0
sint
1
2
---
gt
2
–+=
xv
0
cost=
t
x
v
0
cos
-----------------
=
yy
0
x 
gx
2
2v
o
2

2
cos
-----------------------
–tan+=
dy
d
------
x 
2
gx
2

2
sectan
v
0
2
----------------------------------
–sec 0==

max
tan
v
0
gx
------
2
=