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Projec'le Mo'on
Name: ____________________________
If you throw a ball straight up, its velocity decreases due to a constant downward acceleration (g
= -9.8 m/s
2 assuming upward direction to be positive and negligible air resistance), its velocity
becomes zero at the maximum height and then its velocity becomes negative (since it moves
downward) and keeps increasing in the negative direction until it reaches ground. The velocity
and acceleration graphs as a function of time are shown below.
The equations of motion for an object moving along a straight line (one dimensional motion)
with constant acceleration in this case would be:
If you throw the ball at an angle ball, it exhibits projectile motion.
v f
= v i
y f
= y i
t +
at
2
v
2
f
− v
2
i
= 2 a (
y f
− y i)
Since the ball moves both horizontally (along x-axis) and vertically (along y-axis), we will need
6 equations motion, 3 for the horizontal motion and 3 for the vertical motion. The velocity will
have x and y components, similarly acceleration will also have x and y components.
For motion along x-axis:
And for motion along y-axis:
We will try to perform a few experimental activities, in order to understand the projectile motion
using these equatios.
Open the simulation:
h"ps://phet.colorado.edu/sims/html/projec 4 le-mo 4 on/latest/projec 4 le-mo 4 on_en.html
Double click on the “Vectors” box.
v fx
= v ix
t
x f
= x i
t +
a x
t
2
v
2
fx
− v
2
ix
= 2 a x(
x f
− x )
v f y
= v iy
t
y f
= y i
t +
a y
t
2
v
2
f y
− v
2
iy
= 2 a y(
y f
− y i)
Step 5: Uncheck the “Force Vectors”, and check the “Acceleration Vectors” box. Also check the
“Air Resistance” box to turn on the air resistance. Shoot the ball and notice how acceleration
arrow is pointing down to the left. If it is not obvious, increase the speed to maximum and shoot
the ball again, you will see it. Without the air resistance, the acceleration was pointing straight
down? Why does the air resistance make it tilt to the left? Explain in terms of forces.
The air resistance makes it tilt to the left because of the forces acting against it. The forces act
opposite to the opposite flow velocity. The object its going to tilt to a side or slow down but in
this example it tilted to the left
Step 6: From steps 2 and 3, it should be obvious to you that the horizontal component of the
velocity stays constant but the vertical component of the velocity behaves exactly like that of a
ball thrown straight up, that we discussed earlier (in negligible air resistance). This is because of
the fact that the only force that is causing downward acceleration is gravity, and there is no force
in the horizontal direction, hence there is no horizontal acceleration and thus the horizontal
component of velocity stays constant. So the equations of motion for projectile motion from page
will become:
For motion along x-axis:
And for motion along y-axis:
The first and the third equation for motion along x-axis mean the same thing, i.e. x-component of
the velocity stays constant. So, in fact we have only one equation of motion along x-axis and 3
equations for motion along y-axis. They are:
(1)
(3)
Where: ay = - 9.8 m/s
2 .
Step 7: Let’s see how these equations work.
Assume that a projectile is shot horizontally at 20 m/s from a height of 10 m. That means:
Initial height: yi = 10 m
v fx
= v ix
x f
= x i
t
v
2
fx
− v
2
ix
v f y
= v iy
t
y f
= y i
t +
a y
t
2
v
2
f y
− v
2
iy
= 2 a y(
y f
− y i)
x f
− x i
= v ix
t
v f y
= v iy
t ( 2 )
y f
= y i
t +
a y
t
2
v
2
f y
− v
2
iy
= 2 a y(
y f
− y i)
Step 9: Now use substitute this time t into equation (1) to find out how far the ball hits the
ground (horizontally), from the point of launch. Show all your work below.
xf-xi=(20)(1.43)
=28.6m
Step 10: Let’s test these predictions experimentally. Click on the “Intro” tab at the bottom of the
simulation. This will open the part of the simulation where the cannon is placed on a tower 10 m
above the ground. From the drop-down menu on the top right-hand corner, select “Cannonball”.
Adjust the initial speed to 20 m/s. Drag the target mat to the right until it is exactly at the
distance that you calculated in step 9 above. Then, shoot the ball.
Did the ball hit the target?
Your answer: Yes
Step 11: Drag the “Time,Range.Height” meter and place its crosshair exactly at the
location where the ball hit the ground. The meter should display a height of 0 m.
Take a screenshot of your simulation along with the meter and paste it in the space below.
x f
− x i
= v ix
t
x f
− x i
What is the time reading on the meter?
Your answer: 1.
Does your time reading on the meter match with your prediction in step 8?
Your answer: yes
What is the range reading on the meter (this is how far the ball lands from the point of launch)?
Your answer: 28.
Does your range reading match with your prediction in step 9?
Your answer: yes
Step 12: If a projectile is launched from the ground with a speed of 18 m/s at an angle of 60
o .
a) Calculate the time it spends in the air? Show all your work below.
18sin60=15.6m/s
