Catapult Worksheet
Projectile Motion
1. Consider throwing a ball into the air (with some angle, not straight up), what would the path of the
ball be? Sketch your idea on the following graph, label where the ball hits its maximum height and
maximum distance from the starting location:
Students should graph a parabola or parabolic like shape:
Discuss the different shape graphs and ask students
to justify their graphs with physical explanations.
Extensions:

Should the graph start at the origin?

What factors might change the shape of the
graph? (Prompt them to think about
size/mass of object, air resistance, gravity,
initial speed)
For the following questions we will need to use a few formulas about projectile motion. An object is
considered a projectile behaving under projectile motion rules when it starts with an initial speed and
initial angle and has only the force of gravity and air resistance working upon it to bring it back to the
ground.
Definition of variables:
v0  initial speed of object at time of launch
g  9.8 m s2  32 ft s2 = gravity
  catapult launch angle
The following three variable t, h, and d are instantaneous values at time t:
t  time elapsed from launch
h  height of ball at time t
d = horizontal distance of ball at time t
The following three variables T, H, D are total or maximum values:
T  total time of flight (from launch to landing)
H  maximum height achieved by ball
D  total horizontal distance achieved by ball (at landing)
Projectile Motion Formulas:
To find launch angle
H tan 

D
4
:
 4H 
   tan 1 

 D 
To find initial speed
v0 
v0 of object:
gD
sin 2
2. A soccer ball is kicked into the air, it reaches a maximum height of 3 meters and lands on the
ground 12 meters from where it was kicked:
a. Use the formula given above and your scientific calculator to determine the launch angle  .
Show all your work mathematically in the space provided and then put your answer in the
box provided.
Using the formula given above:
H tan 

D
4
 4H 
   tan 1 

 D 
 4  3m  
1  12m 
1
  tan 
  tan 1  45
 12m 
 12m 
  tan 1 
  45
b. Use the launch angle
 , the formula above and your scientific calculator to find the initial
speed of your ball as it leaves the catapult,
v0 . Show all your work mathematically in the
space provided and then put your answer in the box provided.
Using the formula given above:
v0 
9.8 m s2 12m 

sin 2  45


v0 
gD
sin 2
117.6 m
2
sin  90
s2


117.6 m
1
2
s2
 10.84 m s
v0  10.84 m s  35.4 ft s
3. A ball is launched into the air from a cannon, it reaches a maximum height of 120 feet and lands
on the ground 1525 feet from where it was launched:
Note: The units in this problem are given in US Customary Units (in feet) rather than in International
Standard (SI) Units (metric). You could add a quick lesson on conversions factors at this step.
1525 feet  464.82 m
120 feet  36.576 m
a. Use the formula given above and your scientific calculator to determine the launch angle  .
Show all your work mathematically in the space provided and then put your answer in the
box provided.
Using the formula given above:
H tan 

D
4
 4H 
   tan 1 

 D 
 4 120 ft  
1  480 ft 
1
  tan 
  tan  0.3148  17.47
 1525 ft 
 1525 ft 
  tan 1 
  17.47
b. Use the launch angle
 , the formula above and your scientific calculator to find the initial
speed of your ball as it leaves the catapult,
v0 . Show all your work mathematically in the
space provided and then put your answer in the box provided.
Using the formula given above:
v0 
32 ft s 2 1525 ft 

sin 2 17.47

v0 

gD
sin 2
48800 ft
2
s2
sin  34.94

48800 ft

.5727
2
s2
 85210.4 ft
2
s2
 291.9 ft s
v0  291.9 ft s
If you would like to convert into SI units (or did from the beginning, the result would be:
v0  291.9 ft s  88.97 m s
Place and securely hold the catapult on a level surface and launch the ball, note where it lands. Take
the height measuring stick and place it at the halfway point between the launch and landing. (You will
need to measure the maximum height achieved for the next several launches.)
4. Holding the catapult steady (do not make any changes to the settings), launch the ball several
times and record the maximum height, maximum distance and time of flight. Once you have
collected data from 3-6 launches, average the data at the bottom of each column.
The table is filled in with data gathered in a specific experiment. Students’ data should look
similar when they measure the calculations. The data collected was measure in US customary
units (feet) but the equivalent SI units (meters) are provided for the averages.
Launch
Maximum Height
Maximum Distance
Time of Flight
1
11.5’
93’
1.83
2
13’
90’
1.70
3
12.5’
85’
1.90
4
12’
87’
1.65
5
12.5’
91’
1.73
6
13.5’
89’
1.60
Average
76
 12.5'
6
H  12.5 ft  3.81 m
H
535
 89.2 '
6
D  89.2 ft  27.19m
D
T
10.4 s
 1.73 s
6
5. Use the formula given above and your scientific calculator to determine the launch angle  . Show
all your work mathematically in the space provided and then put your answer in the box provided.
Using the formula given above:
H tan 

D
4
 4H 
   tan 1 

 D 
 4 12.5 ft  
1  50 ft 
1
  tan 
  tan  0.5605  29.3
 89.2 ft 
 89.2 ft 
  tan 1 
  29.3
6. Use the launch angle
 , the formula above and your scientific calculator to find the initial speed of
your ball as it leaves the catapult,
v0 . Show all your work mathematically in the space provided
and then put your answer in the box provided.
v0 
Using the formula given above:
v0 
32 ft s 2  89.2 ft 

sin 2  29.3


gD
sin 2
2854.4 ft
2
sin  58.6
s2

2854.4 ft

.854
2
s2
 3344.15 ft
2
s2
 57.8 ft s
v0  57.8 ft s
v0  57.8 ft s  17.6 m s
Knowing the launch angle and the initial speed of a projectile object, we can make a graph of the path of
the projectile through the air using the following formulas to determine the x and y coordinates at
different times along the path:
x  horizontal coordinate of the ordered pair on the graph
y  vertical coordinate of the ordered pair on the graph
x   v0 cos  t and y  
g 2
t   v0 sin   t
2
Sometimes a little rearrangement of the formulas makes the computation easier:
x   v0 cos  t and y   v0 sin   t 
g 2
t
2
For the next problem we are going to use these formulas along with the values we found for the launch
angle
  29.3
x   57.8cos  29.3  t  50.4t
x  50.4t
and the initial velocity
and
v0  57.8 ft s :
y   57.8sin  29.3  t 
 9.8 t 2  28.3t  4.9t 2
2
y  28.3t  4.9t 2
7. Divide your total time from the last column of collected data above by 10. This will be your
increment of time. Fill in the table below in increments starting at 0 and increasing by the
increment determined.
To find the increment of time:
Elapsed
Time
t
1.73 s
 .173 s
10
x – Coordinate
y-coordinate
x
y
0
0
 0, 0 
.173 s
8.72
4.41
.346 s
17.4
7.87
.519 s
26.2
10.37
.692 s
34.9
11.91
.865 s
43.6
12.46
1.038 s
52.3
12.16
1.211 s
61.0
10.8
1.384 s
69.8
8.55
1.557 s
78.6
5.2
1.73 s
87.2
1
8.72, 4.41
17.4,7.87 
 26.2, 10.37 
34.9, 11.91
 43.9, 12.46
52.3, 12.16
 61.0, 10.8
 69.8, 8.55
 78.6, 5.2
87.2, 1
0
8. Plot your ordered pairs
 x, y 
(x, y)
on the graph below; adjust the scale on your axes as necessary.
Below is an Excel generated graph of the points found in the above table. Students should have a
parabolic shaped graph after they plot the points in their table.
9. Connect the points you graphed.
Was your earlier graphical prediction
about projectile motion correct?
What type of a curve is this?
10. In order to find the launch angle of the catapult arm that will maximize the distance covered by the
ball, you will need to manipulate the formula
v0 
gD
and solve for D. Show your work
sin 2
mathematically and then place the formula for D in the box given.
Start with
v0 
gD
sin 2
Square both sides to eliminate the square root:
 gD
 v0   
 sin 2
gD
v0 2 
sin 2
2



2
Now multiply both sides by
v0 2  sin 2  
sin  2  :
gD
 sin 2 
sin 2
v0 2  sin 2   gD
Divide both sides by g to isolate D:
v0 2  sin 2  gD

g
g
v0 2  sin 2 
D
g
D
v0 2  sin 2 
g
11. Use the initial launch velocity
vo found in #7, the formula you found for D in #11, and your
scientific calculator to find the horizontal distance D the ball will travel for the following launch
angles of the catapult arm.
v0 2  sin 2 
Using the formula: D 
g
57.82  sin 2  3340.84 sin 2 
D

9.8
32
D  104.4  sin 2 
Launch Angle

Distance
D
5
18.1 ft
15
52.2 ft
25
80 ft
35
98.1 ft
45
104.4 ft
55
98.1 ft
65
80 ft
75
52.2 ft
85
18.1 ft
12. Based off your calculations in the table above, what would be the best launch angle  for the
catapult arm if you want to get the ball to travel the maximum possible distance from the catapult?
At
45 the distance is maximized at 104.4 ft.
13. If you set the catapult arm at this angle
At
 , how far will the ball land from the catapult?
45 the distance is maximized at 104.4 ft.
14. Now you need to adjust the catapult base so the launch angle will give you a maximum distance.
You will need to change the launch angle from the initial launch angle found in #6 to the angle
 you determined would net the maximum distance for the ball, found in #13. Determine the
adjustment factor between your two angles and place your answer in the box. For example, if you
found the initial launch angle in #6 was
36 and the angle that would net the maximum distance in
#13 to be 52 then your catapult will need to be adjusted by 52  36  16 .
The test catapult had a launch angle of
29.3  30 .
Sind the angle that maximizes the distance is
adjustment factor is found:
Adjustment Factor  15
45 , the
45  30  15
15. In order to launch the ball at the angle that will achieve the maximum distance, you will need to
create an angle with the floor and the catapult base by placing wooden blocks under the front of
the catapult. Carefully measure the angle between the catapult and the floor until you have raised
the base up to the adjustment factor found in the last step. Hold the catapult securely at the
correct angle. Fire the catapult and record the distance. This should be longer than for an
unadjusted launch. Record this distance.
After students have adjusted their catapults, they should
find the distance of the launch to be close to 104 ft.
Distance = 104 ft
16. Suppose you want to launch the ball from the catapult to hit a target that lies at a point set at 2/3
of the maximum distance found above. Use the formula given below and your scientific calculator
to determine to determine the angle required to achieve this distance. Show all your work
mathematically in the space provided and then put your answer in the box provided.
1
 gD 
  sin 1  2 
2
 v0 
2
of the maximum distance of 104 ft is:
3
2
104 ft   69.3 ft
3
Using the formula
 gD 
1
  sin 1  2  :
2
 v0 
ft
1 1   32 s2   69.3 ft  


  sin
  57.8 ft s 2 
2


 2218 ft


ft
 3314 s2 
1
2
  sin 1 
2
s2
2
1
2
  sin 1 .669 
  21.0
  21.0
17. Set the catapult to launch at the angle determined for this distance. The catapult base will again
have to be adjusted to achieve this requirement. Place wooden blocks appropriately until you have
adjusted the launch angle to the proper degree measurement and hold securely at the correct
angle. Fire the catapult and record the distance. The distance should be close to desired.
After students have adjusted their catapults,
they should find the distance of the launch to be close to 70 ft.