PROJECTILE
MOTION
WHAT IS A PROJECTILE?
• An object designed to be fired from a cannon.
WHAT IS PROJECTILE MOTION?
• When objects follow a curved path because it is moving
horizontally and vertically simultaneously.
• The vertical motion is analyzed like a falling object.
• The horizontal motion follows Newton’s 1st Law (constant
motion in a straight line).
2 TYPES OF PROJECTILES
Objects launched at
an angle
Objects horizontally
launched
HORIZONTALLY LAUNCHED
OBJECTS
HORIZONTALLY LAUNCHED
• Motion
happens in 2 dimensions:
20 m/s
OBJECTS
horizontal and vertical
• Initial vertical velocity = 0 but
increases as object falls due to
gravitational forces. So the
vertical velocity changes.
• Initial horizontal velocity is NOT
ZERO and remains unchanged for
the duration of the flight because
there are NO horizontal forces.
(Newton’s 1st Law)
20 m/s
20 m/s
20 m/s
HORIZONTALLY LAUNCHED
20 m/s
OBJECTS
Horizontal Velocity
20 m/s
20 m/s
20 m/s
20 m/s
10 m/s
HORIZONTALLY LAUNCHED
OBJECTS
Vertical Velocity
-9.8 m/s
-9.8 m/s
Click to add text
-19.6 m/s
-19.6 m/s
-29.4 m/s
-29.4 m/s
Remember slope on velocity time
graph = acceleration.
HORIZONTALLY LAUNCHED OBJECTS
Vertical Acceleration
-9.8 m/s
-9.8 m/s
-19.6 m/s
-9.8
-9.8 m/s
-29.4 m/s
OBJECTS LAUNCHED AT AN ANGLE
final
initial
OBJECTS LAUNCHED AT AN ANGLE
• Motion occurs in 2 dimensions: horizontal
and vertical
• Initial vertical velocity is NOT ZERO and
changes during the journey because of the
gravitational force
• Initial horizontal velocity is NOT ZERO and
remains unchanged for the duration of the
flight because there are NO horizontal forces.
(Newton’s 1st Law)
• Time and speed are symmetrical as long as
the launch height and landing position are
the same.
OBJECTS LAUNCHED AT AN ANGLE
30 m/s
Horizontal Velocity
30 m/s
30 m/s
30 m/s
30 m/s
30 m/s
20
m/s
10
m/s
30 m/s
30 m/s
OBJECTS LAUNCHED AT AN ANGLE
0 m/s
9.8 m/s
Vertical Velocity
-9.8 m/s
30 m/s
29.4 m/s
19.6 m/s
19.6 m/s
9.8 m/s
-19.6 m/s
0 m/s
-9.8 m/s
-19.6 m/s
-30 m/s
-29.4 m/s
29.4 m/s
-29.4 m/s
OBJECTS LAUNCHED
AT AN ANGLE
Vertical Acceleration
0 m/s
-9.8 m/s
-9.8 m/s
9.8 m/s
-9.8 m/s
-9.8 m/s
-9.8 m/s
19.6 m/s
-9.8 m/s
-19.6 m/s
-9.8 m/s
-9.8 m/s
29.4 m/s
-29.4 m/s
OBJECTS LAUNCHED AT AN
IfANGLE
an object is launched at the angles below with the same
initial velocity, which one will land the furthest away?
Answer: 45 degrees
OBJECTS LAUNCHED AT AN
ANGLE
What pairs of angles will have the same range for
the projectile?
Answer: 15 and 75 degrees & 30 and 60 degrees
OBJECTS LAUNCHED AT AN
ANGLE
Which angle would keep the projectile in the air
for the longest period?
OBJECTS LAUNCHED AT AN
ANGLE
Which
angle has the maximum height for the projectile
to travel?
PROJECTILE
CALCULATIONS
KINEMATIC EQUATIONS
A STONE IS THROWN HORIZONTALLY FROM THE TOP OF A
CLIFF WITH A VELOCITY OF 15 M/S. IT TAKES 3 S FOR
THE ROCK TO LAND.
A) HOW HIGH IS THE CLIFF?
Given
Unknown
Equation
Substitute
Solve
A STONE IS THROWN HORIZONTALLY FROM THE TOP OF A
CLIFF WITH A VELOCITY OF 15 M/S. IT TAKES 3 S FOR
THE ROCK TO LAND.
A) HOW HIGH IS THE CLIFF?
Given
Vi = 15 m/s
T = 3s
Unknown
Equation & Substitute
Height (Yf) Yf – Yi = Vi*t – ½ g t^2
89.1 m
Yf – 0 = (15m/s)(3s) - ½ (-9.8m/s^2)(3^2)
g= -9.8 m/s^2
Yi = 0
Solve
Yf = 45 m/s^2 - (- 44.1)
Yf = 89.1 m
A STONE IS THROWN HORIZONTALLY FROM THE TOP OF A
CLIFF WITH A VELOCITY OF 15 M/S. IT TAKES 3 S FOR
THE ROCK TO LAND.
B) HOW FAR FROM THE BASE OF THE CLIFF DOES THE
STONE STRIKE THE GROUND?
Given
Unknown
Equation
Substitute
Solve
A STONE IS THROWN HORIZONTALLY FROM THE TOP OF A
CLIFF WITH A VELOCITY OF 15 M/S. IT TAKES 3 S FOR
THE ROCK TO LAND.
B) HOW FAR FROM THE BASE OF THE CLIFF DOES THE
STONE STRIKE THE GROUND?
Given
Unknown
Equation & Substitute
Vi = 15 m/s
T = 3s
Distance (Xf)
Xf – Xi = Vi*t + ½ at^2
0.9 m
Xf – 0 = 15*3 + ½ (-9.8)(3^2)
a=g= -9.8 m/s^2
Xf = 45 –44.1
Yi = 0
Xf = 0.9 m
Xi = 0
Solve
A STONE IS THROWN HORIZONTALLY FROM THE TOP OF A
CLIFF WITH A VELOCITY OF 15 M/S. IT TAKES 3 S FOR
THE ROCK TO LAND.
C) HOW FAST IS IT GOING WHEN IT HITS THE GROUND?
Given
Unknown
Equation
Substitute
Solve
A STONE IS THROWN HORIZONTALLY FROM THE TOP OF A
CLIFF WITH A VELOCITY OF 15 M/S. IT TAKES 3 S FOR
THE ROCK TO LAND.
C) HOW FAST IS IT GOING WHEN IT HITS THE GROUND?
Given​
Unknown​
Equation & Substitute​
Solve​
Vi = 15 m/s​
T = 3s​
Final Velocity (Vf)
Vf^2 = Vi^2 + 2g(Yf – Yi)
41.78 m/s
a=g= -9.8 m/s^2​
Yi = 0
Vf^2 = 0 + 2(-9.8)(89.1 – 0)
Vf^2 = 1,746.3
Xi = 0​
Yf = 89.1 m
Vf =
Xf = 0.9 m
Vf = 41.78 m/s
Vi = 0 in Y direction
1,746.3