The Research Experience for Teachers Program
http://www.cs.appstate.edu/ret
Activity Title: Introduction to Projectile Motion
Introduction/Motivation: The equations used to describe projectile motion are generally derived from
the basic dvat equations for horizontal and vertical motion. More advanced equations using basic right
triangle trigonometry and vectors provide a further analysis of projectile motion. Generally, students
complete one word problem at a time, only using one small group of equations. Students rarely, if ever
have any understanding of how objects undergoing projectile motion can be described in terms of
changing velocity, mass, kinetic energy, gravitational potential energy, total mechanical energy and
momentum during and objects motion. In this activity, you will be able to create basic formulas and
analyze the meaning of graphs that have been previously prepared as part of the spreadsheet.
Learning Objectives: You will be able to:



Correctly input equations into Excel using constants, basic algebra and trigonometry functions
Analyze graphs of projectile motion with concept of angle, range and altitude
Analyze velocity and energy motion diagrams for projectile motion over a range of initial angles
Materials List: You will need one computer per student with Microsoft Excel program
Background: You should have your noted and basic equations regarding projectile motion with you
during this practice.
Pre-Assessment:
1. What happens to the kinetic energy in terms of horizontal velocity of projectiles in motion as the
initial angle increases?
2. What happens to the kinetic energy in terms of the initial vertical velocity of projectiles in
motion as the initial angle increases?
3. At what angle would a projectile have the maximum range?
4. At what angle would a projectile have the maximum height?
5. What relationship exists between the kinetic energy, gravitational energy, and total mechanical
energy of a projectile at any angle?
Lab Activity: Day 1 (1 – 1.5 hours)
Projectile Motion: Basic
1.
Open the spreadsheet labeled “ Projectile Motion.xlsx” in Excel. Start on the page that says
Projectile Motion Basic at the bottom of the page.
2. Notice that all the cells are empty except for the angles in degrees and a section on the far right
in the color blue. Do NOT change anything in the color blue.
Projectile Motion Page 1
3. You will also notice that the graphs below the spreadsheet are blank. As you complete
calculations in the spreadsheet, they will slowly fill in.
4. Make sure that you are beginning on the spreadsheet that is labeled “Projectile Motion – Basic”
at the bottom. You will also notice that cell C5 should already have the number 0 for initial
height, therefore the initial height of each of these projectiles is 0 meters.
5. In cell C9 type the number 10, therefore all of these projectiles will have an initial velocity of
6. 10 m/s. You will be able to change the initial velocity later.
7. In cell C12, change the mass of the projectile to 5, therefor the mass of this projectile is 5 kg.
You will be able to change the mass of the projectile later
8. Goto cell F11, you must convert all of your degrees data into radians. The functions cosine and
sine in Excel are specific to units of radians, and will not give the correct answer if you attempt
to take the cosine and sine of an a number in units of degrees.
9. The equation written in cell F10 is rad=(deg*2pi)/360. You are calculating for an angle in
radians, type the equation going from degrees into radians in cell F. ***Remember that Excel
already has pi as a constant, simply type pi() in the multiplication part of the formula.
10. Copy your equation and results from cell F11, then paste down the column F until cell F101, just
before the graphs, when the angle is 90o. (90o should convert to 1.5708 radians)
11. Goto cell G11, calculate the horizontal velocity. The equation written in cell G10 is Vx =
Vrcos(), ***Remember that Vr, the initial or resultant velocity is a constant, and must be
written as $C$9 in your Excel formula.)
12. Copy your equation and results from cell G11, then paste down the column G until cell G101,
just before the graphs, when the angle is 90o. (Vx at 90o should be 0.000) (Answer questions 1
and 2 below) The top of your spreadsheet should look like this”
initial height
m
0
Initial or Resultant Velocity
m/s
10
Angle
Angle
Degrees Radians
theta

mass of projectile
kg
5
horizontal velocity
m/s
Vx
rad=(deg*2pi)/360
Vx = Vrcos()
0
0.0000
10.0000
1
0.0175
9.9985
2
0.0349
9.9939
3
0.0524
9.9863
4
0.0698
9.9756
5
0.0873
9.9619
13. Goto cell H11, calculate the horizontal momentum (px). The equation written in cell H10 is Px =
m*Vx
***Remember that m, the mass, is constant, and must be written as $C$12 in your Excel
formula.)
Projectile Motion Page 2
14. Goto cell G11, calculate the horizontal kinetic energy(Ex). The equation written in cell G10 is
Ex = (1/2)*m*Vx^2 ***Remember that m, the mass, is constant.
15. Copy your equation and results from cells H11 and I 11, then paste down the column H and I
until cells H101, and I101. (Answer question 3 below)
16. Goto cell J11, calculate the initial vertical velocity from the equation Voy=-Vrsin(), copy and
paste down column J. All of your answers should come out as negative velocities. (Answer
question 4 below.)
17. Determine the initial vertical momentum and kinetic energy in columns K and L. Copy and paste
the formulas and calculations down column K and L until the graphs start. ***Remember that
m, the mass, is constant You will see the formulas for momentum and kinetic energy on the
spreadsheet. (Answer question 5 below.) The top of your spreadsheet should appear as:
initial
initial
horizontal momentum horizontal energy vertical velocity vertical momentum
kg*m/s
Joule = kg*m^2/s^2
m/s
kg*m/s
Px
Ex
Voy
Poy
Px = m*Vx
50.0000
49.9924
49.9695
49.9315
49.8782
49.8097
Ex = (1/2)*m*Vx^2 Voy=-Vrsin()
250.0000
0.0000
249.9239
-0.1745
249.6955
-0.3490
249.3152
-0.5234
248.7835
-0.6976
248.1010
-0.8716
Poy = m*Voy
0.0000
-0.8726
-1.7450
-2.6168
-3.4878
-4.3578
initial
vertical kinetic energy
Joule = kg*m^2/s^2
Ex
Ex = (1/2)*m*Voy^2
0.0000
0.0761
0.3045
0.6848
1.2165
1.8990
18. Goto cell M11, calculate the final vertical velocity of the projectile, copy and paste down column
M to M101, just before the graphs. (Answer question 6 below)
19. Determine the final vertical momentum and kinetic energy in columns K and L. Copy and paste
the formulas and calculations down column N and O until the graphs start. ***Remember that
m, the mass, is constant You will see the formulas for momentum and kinetic energy on the
spreadsheet. (Answer question 5 below.)
20. Goto cell P11, you will the the equation for the resultant momentum Pr in terms of the
Pythagorean theorem, Pr^2 = Px^2 + Pfy^2 , therefore you will have to take the square root of
both sides of the equation to solve for Pr. Reminder, to take the square root in excel, you must
take the power of your whole function to .5. Copy and paste down column P. The value of Pr at
90o should be 70.7107
21. Determine the final kinetic energy in column Q. Copy and paste the formulas and calculations
down column Q. (Answer questions 8 and 9 below.) The top of your cells should appear as:
Projectile Motion Page 3
final
vertical velocity
m/s
Vfy
Vfy=-Voy
0.0000
0.1745
0.3490
0.5234
0.6976
0.8716
final
final
vertical momentum vertical kinetic energy
kg*m/s
Joule = kg*m^2/s^2
Pfy
Efy
Pfy = m*Vfy
0.0000
0.8726
1.7450
2.6168
3.4878
4.3578
Efy = (1/2)*m*Vfy^2
0.0000
0.0761
0.3045
0.6848
1.2165
1.8990
resultant
momentum
kg*m/s
Pr
resultant
kinetic energy
Joule = kg*m^2/s^2
Er
Pr^2 = Px^2 + Pfy^2
50.0000
50.0000
50.0000
50.0000
50.0000
50.0000
Er = Ex + Eoy
250.0000
250.0000
250.0000
250.0000
250.0000
250.0000
22. In columns R and S you will calculate the maximum height of the projectiles using two different
equations. In column R, you will calculate the height based on the final vertical velocity
h= (Vfy^2)/(2*g), while in column S, you will calculate the height from the intitial thrust or
resultant velocity and the angle, h= ((Vr^2)*(sin)^2)/(2*g). As a check, both columns must
have the same result or something is wrong. The value for g, the acceleration of gravity is found
in cell R3, and is a constant so, it will be written as $R$3 in your equation, and the value for Vr in
the second equation will also be a constant, $C$9. Use this cell in the equation because you will
be able to change the acceleration of gravity to other planets later.
23. Goto cell T11 and calculate the gravitational potential energy at the maximum height. Copy and
past down column T. You may use either column for height in your calculation. Both mass and
the acceleration of gravity will be constants in this equation.
24. Goto cell U11, you will calculate the range of the projectile when given the initial thrust or
resultant velocity and the acceleration of gravity, both constants, with an increasing angle as in
the equation R = (Vr^2)(sin(2*)/g. Note that in the equation, it is the sin(2*), NOT the sine
squared. (Answer questions 10, 11 and 12 below.)
25. In columns V and W, you will calculate the total time in the air using two different equations
based on different given information. To check, both columns will calculate the same result or
something is wrong.
26. Finally, you will determine the maximum range in column X using the horizontal velocity and
either column of time you just calculated. As a check, column X and column U must give the
same results.
27. Look at the Energies of Projectile vs. Angle and the Trajectory Motion for Initial Thrust Vr graphs
and answer questions 13 – 16 below.
Projectile Motion Page 4
max
height
m
h
max
height
m
h
potential energy
at max height
Joule = kg*m^2/s^2
GPE
Maximum
Range
m
R
Total
time
t
s
Total
time
t
s
Maximum
Range
R
m
h= (Vfy^2)/(2*g)
0.0000
0.0016
0.0062
0.0140
0.0248
0.0388
h= ((Vr^2)*(sin)^2)/(2*g)
0.0000
0.0016
0.0062
0.0140
0.0248
0.0388
GPE = mgh
0.0000
0.0761
0.3045
0.6848
1.2165
1.8990
R = (Vr^2)(sin2*)/g
0.0000
0.3561
0.7118
1.0666
1.4201
1.7719
t = (2Vfy)/g)
0.0000
0.0356
0.0712
0.1068
0.1424
0.1779
t = 2(2h/g)^.5
0.0000
0.0356
0.0712
0.1068
0.1424
0.1779
R = Vx*t
0.0000
0.3561
0.7118
1.0666
1.4201
1.7719
Lab Activity: Day 2 (1 – 1.5 hours)
Projectile Motion – Advanced
1.
2.
3.
4.
5.
Open the spreadsheet labeled “ Projectile Motion.xlsx” in Excel. Start on the page that
says Projectile Motion-Advanced at the bottom of the page.
Notice that all the cells are empty except for the angles in degrees and a section on the far
right in the color blue. Do NOT change anything in the color blue.
You will also notice that the graphs below the spreadsheet are blank. As you complete
calculations in the spreadsheet, they will slowly fill in.
Make sure that you are beginning on the spreadsheet that is labeled “Projectile Motion –
Advanced” at the bottom. Unlike the original projectile motion page where projectiles are
shot from the surface of flat ground, on the advanced page, the projectiles are shot from a
known height, such a cliff.
Start with a height of 35 in cell C5, an initial velocity of 20 in cell C9, and a mass of 15 in cell
C12. Notice the units for each is in the cell above the number. Each of these will be
constants in the upcoming equations. Cells should look like this:
initial height (h2)
m
10
Initial or Resultant Velocity
m/s
10
mass of projectile
kg
0.15
Angle
Angle
Degrees Radians
theta

0
1
2
3
4
5
rad=(deg*2pi)/360
0.0000
0.0175
0.0349
0.0524
0.0698
0.0873
horizontal velocity horizontal momentum
m/s
kg*m/s
Vx
Px
Vx = Vrcos()
10.0000
9.9985
9.9939
9.9863
9.9756
9.9619
Px = m*Vx
1.5000
1.4998
1.4991
1.4979
1.4963
1.4943
6. Columns F through N follow the same directions as the Projectile Motion- Basic table, so
convert the angle from degrees to radians, and compute all the way through the initial total
vertical mechanical energy. Copy and paste data/calculations for columns F through N.
Answer question 17 below. Cells should look like this:
Projectile Motion Page 5
initial
initial
horizontal energy vertical velocity vertical momentum
Joule = kg*m^2/s^2
m/s
kg*m/s
Ex
Voy
Poy
Ex = (1/2)*m*Vx^2 Voy=-Vrsin()
7.5000
0.0000
7.4977
-0.1745
7.4909
-0.3490
7.4795
-0.5234
7.4635
-0.6976
7.4430
-0.8716
Poy = m*Voy
0.0000
-0.0262
-0.0523
-0.0785
-0.1046
-0.1307
initial
initial vertical
initial total
vertical kinetic energy gravitational potential energy vertical mechanical energy
Joule = kg*m^2/s^2
Joule = kg*m^2/s^2
Joule = kg*m^2/s^2
Ey
PE
Etoty
Ey = (1/2)*m*Voy^2
0.0000
0.0023
0.0091
0.0205
0.0365
0.0570
PE= mgh (use h2)
14.7000
14.7000
14.7000
14.7000
14.7000
14.7000
Etoty = Ey + PE
14.7000
14.7023
14.7091
14.7205
14.7365
14.7570
7.
In cell O11, you will now calculate the final vertical velocity of the projectile using the
equation Vfy = ((-Voy)^2 + 2gh)^.5 Both g and h are constants in this equation. Answer
question 18 below. Copy and paste down for column O.
8. Columns P through S follow the same directions as the Projectile Motion- Basic table, so
calculate the final vertical momentum, and compute all the way through the resultant
kinetic energy. Copy and paste data/calculations for columns P through S. Answer question
19 below. Cells should look like this:
final
vertical velocity
m/s
Vfy
(use h2)
Vfy = ((-Voy)^2 + 2gh)^.5
14.0000
14.0011
14.0043
14.0098
14.0174
14.0271
final
final
vertical momentum vertical kinetic energy
kg*m/s
Joule = kg*m^2/s^2
Pfy
Efy
Pfy = m*Vfy
2.1000
2.1002
2.1007
2.1015
2.1026
2.1041
Efy = (1/2)*m*Vfy^2
14.7000
14.7023
14.7091
14.7205
14.7365
14.7570
vector
resultant
momentum
kg*m/s
Pr
scalar
resultant
kinetic energy
Joule = kg*m^2/s^2
Er
Pr^2 = Px^2 + Pfy^2
2.5807
2.5807
2.5807
2.5807
2.5807
2.5807
Er = Ex + Eoy
22.2000
22.2000
22.2000
22.2000
22.2000
22.2000
9. In cell T11, you will calculate the maximum height to the new apex. You will use the
equation htot = h1 + h2 = ((Vr^2)*(sin)^2)/(2*g) +h2, where h1 = ((Vr^2)*(sin)^2)/(2*g).
htot stands for the total height. h1 is the altitude from the top of h2 to the apex. Vr, h2 and
g are all constants. Copy and paste down column T.
10. In cell U11, calculate the total mechanical energy at the maximum altitude. Remember,
htot = h1+h2. This htot was calculated in column T. Copy and paste down column U.
11. In cells V and W, use the given equations to calculate the time to the sapex, and the time
from the apex to the ground. In column X you will calculate the range. In column Y
calculate the impact velocity Vri with the ground, and in column Z, you will determine the
angle (in radians) upon impact. Finally, in column AA convert radians back to degrees.
Answer question 20 below: Cells should look like:
Assessment:
Projectile Motion Page 6
g on earth
9.8
maximum height
at new apex
m
h1 = ((Vr^2)*(sin)^2)
htot = h1 + h2
total
potential energy
at max height
Joule = kg*m^2/s^2
GPE
htot= ((Vr^2)*(sin)^2)/(2*g) +h2
10.0000
10.0016
10.0062
10.0140
10.0248
10.0388
GPE = mg(h1+h2)
14.7000
14.7023
14.7091
14.7205
14.7365
14.7570
Time to apex
t1
s
Time from apex
to ground
t2
s
t1 = ((2*Vr^2*sin2)/g^2) t2 = 2(2htot/g)^.5
0.0000
2.8571
0.0727
2.8574
0.1453
2.8580
0.2177
2.8591
0.2898
2.8607
0.3616
2.8627
Maximum
Range
R
m
R = Vx*(t1+t2)
28.5714
29.2960
30.0147
30.7260
31.4284
32.1202
Final Resultant
Impact Velocity
Vri
m/s
on excel, arctan
is = atan
Impact Angle
radians
Impact Angle
degrees
Vri=(Vx^2 + Vfy^2)^.5 =tan-1(Vfy/Vx) deg=(360xrad)/(2pi)
17.2047
0.9505
54.4623
17.2047
0.9507
54.4686
17.2047
0.9510
54.4872
17.2047
0.9515
54.5184
17.2047
0.9523
54.5620
17.2047
0.9533
54.6179
Questions: Projectile Motion Basic: Give all responses in terms of physics principles.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
As the angle increases from 0o to 90o, what occurs to the horizontal component of the velocity?
How is the relationship shown in the first graph of Horizontal Velocity vs. Angle?
Why is both the momentum and kinetic energy for horizontal velocity equal to zero at 90o?
Why are all the initial vertical velocities negative?
While the initial vertical momentum is negative, why will the initial kinetic energy never be
negative?
What relationship exists between the horizontal velocity and the final vertical velocity of the
projectile, and at what angle are they equal to each other?
What relationships exist between the initial and final vertical momentum and kinetic energies
for a projectile?
What is significant about the answers for the final total kinetic energy of the projectile?
Energy is a scalar, while momentum is a vector, what is the difference between a scalar and a
vector and why do we find the total of each one using a different equation?
At what angle does the range have the largest value?
What occurs to the values of the range before and after the largest value?
According to the graph below, when is it possible to have a maximum height greater than the
range for a projectile? Why?
What relationship exists between the final vertical kinetic energy and the gravitational potential
energy of the projectile?
Write a mathematical expression that shows the relationship between the total kinetic energy,
the horizontal kinetic energy, and the final vertical kinetic energy. How does this describe a
scalar?
Write a mathematical expression that shows the relationship between the total kinetic energy,
the horizontal kinetic energy, and the gravitational potential energy. How does this describe a
scalar?
Fill in the blanks in regards to a projectile: Projectiles shot from angles that have
complementary angles will have the same __________________, but not the same
____________________ except at the angle of _____ degrees.
Questions: Projectile Motion Advanced: Give all responses in terms of physics principles.
17. Which values have the same value for all angles and why?
Projectile Motion Page 7
18. Why do the values of the final vertical velocity continue to remain greater than the magnitude
of the initial vertical velocity and/or the initial horizontal velocity. (Remember in the Projectile
Motion Basic sheet the initial and final vertical velocities are of equal magnitude but opposite
direction.)
19. Why does the resultant momentum remain the same?
20. How do the two trajectory graphs contrast and compare?
References: Spreadsheet for work is here:
My Module Work\Projectile Motion.xlsx
Projectile Motion Page 8