Tutorial for Projectile Motion
Problems
Components of Velocity
If an object is launched at an angle,
its velocity is not perfectly horizontal
(x direction), nor is it perfectly
vertical (y direction).
An object launched at an angle will
have a velocity that is a combination
of the Vx and the Vy.
You must always find the
components of the initial velocity.
If the velocity is given, you must use
the angle to determine what part of
the initial velocity is in the X
direction and what part is in the Y
direction.
Find V1y and Vx
V0y=V0sin Θ (this is the equation
for V0y always)
Vx=V0 cos Θ
These are known as the
“components” of the velocity.
Changes in Velocity
The velocity in the Y direction must
change. The velocity in the Y
direction decreases because any
object that goes up must be affected
by gravity.
The velocity in the X direction does
not change at all, it is the same
EVERYWHERE along the path.
thalf 
v2 y  v1 y
g
Looking for half time
The time it takes for an object to reach
its maximum height will be half of the
time it spends in the air.
At the maximum height, the velocity in
the Y direction will be 0 m/s
thalf 
v1 y  v0 y
g
Total Time
The total time an object is in the air
is simply twice the half time.
The total time is used to find the
maximum range (X) of the object
(measured in meters).
Finding Max Height
To find maximum height, use the following
equation:
The time in this equation must be the half
time!
1 2
y  v0 y t  gt
2
To Find Range
To find the maximum range, you must use
the total time in the following equation:
x  vx t
Notes about solving problems
If the equation calls for V0y, you
must use the V0y value you
calculated using V0sinΘ.
If the equation calls for Vx, you must
use the Vx value you calculated
using V0cos Θ.
Please note that Vx, V0y, and V0 are
not the same value!!!
Finding the V1 at a specific time
If you are asked to find the V1 at a
specific time, you must first find the
components of the initial velocity (Vx and
V0y).
Next, find the V1y at the given time using
the below equation:
v1 y  v0 y  gt
Finding V1 continued
Once you have V1y at the given time,
and you have Vx, simply use the
Pythagorean Theorem to find V1:
v1  vx  v1 y
2
2
Finding the Angle at Any point
To find the angle at any point, first find
the Vx and the Vy at that point.
Once those are determined, use the
tangent trig function.
Θ=Tan-1 (Vy/Vx)
Example Problems
1.
A ball is kicked into the air at 30 m/s at an angle
of 25 degrees. What is its maximum range and
height?
Given:
V1: 30
m/s
Θ=25
degrees
V2y=0
m/s
g= -9.8
m/s2
Vx=V0 cos Θ=30 cos 25= 27.19 m/s
V0y=V0 sin Θ= 30 sin 25=12.68 m/s
Find:
Vx
V1y
th
tt
X
Y
thalf 
1 2
y  v0 y t  gt
2
Ttotal=2.59 s
v1 y  v0 y
g
thalf 
0  12.68
 1.29s
9.8
1
y  (12.68)(1.29)  (9.8)(1.292 )  8.20m
2
x  vx t
x  (27.19)(2.59)  70.36m
Example Problems
A rocket is launched from the ground and is aimed 2500 m
away. If it reaches a maximum height of 200 meters, how
fast must it be going when it leaves the ground (find Vy and
Vx)? At what angle must it be launched? (Hint: find time
first using cliff problems)
Given:
X: 2500 m
Y: 200 m
Find:
V1y
Vx
th
tt
V1
To find the initial velocity (V0), find Vx and V0y.
First find time using the initial velocity in the y
direction as zero.
T Total=12.78s
t 
V0y= V1y-gt
= 62.62 m/s
v0  vx  v0 y
2
2
y
1/ 2 g
= 6.39s
Vx= x = 195.62 m/s
t
v0  195.622  62.622  205.40m / s
3. A basketball player jumps up at 15
m/s at an angle of 50 degrees. At
what time will he be 2 meters away
(horizontally) from where he
jumped?
Given:
V0: 15 m/s
Θ=50 degrees
X=2 m/s
Find:
t
Vx=V0 cos Θ=9.64 m/s
T=x= .21 s
Vx
4. A football is kicked with an initial
velocity of 55 m/s at an angle of 60
degrees. At what times will it pass
5 meters above the ground? (you
can use quadratic formula in the
calculator)
V0y= V0 sin Θ= 47.63 m/s
Given:
V0: 55 m/s
Θ=60 degrees
Y=5 m
Θ
Find:
T1
T2
T1=.11s
T2= 9.61s
The ball will pass 2 meters high
twice.
You can use the quadratic formula
to find the two times
ax2+ bx + c=0
Make equation look like this
1/2gt2+V1yt-y=0
5. A baseball is thrown into the air with a velocity of
27.5 m/s at an angle of 45 degrees. What will its
velocity be after 3 seconds? (Hint: find its Vy
and Vx at 3 seconds, use the Pythagorean
Theorem).
Given:
V0: 27.5 m/s
Θ=45 degrees
T= 15 sec.
Find:
V1
V1y
Vx
V0y
v1  vx  v1 y
2
Vy= V0 sin Θ=19.45 m/s
Vx= V0 cos Θ= 19.45 m/s
V1y=V0y+gt=9.65 m/s
2
v1  19.452  .9.652  21.71m / s
6. A rocket is shot up into the air. If it
has a range of 1000 m, and reaches
a height of 800 m, what angle was
it shot up at and what was its initial
velocity?
Givens:
X=1000 m
Y= 800 m
G=9.8 m/s2
You must first find the half way time, how long
Find: it the rocket to drop the 800 m; set V1y equal
to zero for this.
V0
800
V0y
t
 12.78s
Vx
1/ 2(9.8)
th
tt
Next, find the total time by doubling the half way time
Ttotal= 25.56 seconds
y
t
1/ 2 g
Find Vx using the given distance (x) and the total time
x
vx 
t
1000
vx 
 39.13m / s
25.56
#6 Continued
Find V1y by setting V2y equal to zero and using the half way
time
v0 y  v1 y  gt  0  (9.8)(12.78)  125.24m / s
Find the V1 by using the Pythagorean Theorem with V1y and Vx.
v0  vx 2  v0 y 2
v0  39.132  125.242  131.21m / s
Find the angle that it is launched at by using the tangent function
  tan (
1
V0 y
Vx
)
125.24
  tan (
)  72.65deg
39.13
1
7. At what times will a rocket launched at an angle
of 37 degrees and a velocity of 400 m/s pass
100 m high? Use Quadratic formula for this.
Given:
Θ=37 degrees
V0= 400 m/s
Y=100 m
Find:
t1
t2
V1y
Find V1y first:
v0 y  v0 sin 
v0 y  400(sin 37)  240.73m / s
To find the two times that the rocket passes that height, you must
use the Quadratic formula: Set y=V0yt+1/2 gt2 to look like ax2+
bx+c=0. You are solving for t, so your A value is -4.9, your B value
is 240.73 m/s (V0y), and your C value is -100 . Use the calculator to
plug in and find the two times it passes those two heights.
T1= .42 s
t2= 48.71 s