Projectile motion refers to the two-dimensional motion of objects through both horizontal and vertical components. There are three types of projectile motion: horizontal, vertical, and parabolic. Horizontal motion involves constant horizontal velocity, vertical motion involves acceleration due to gravity with changing vertical velocity, and parabolic motion follows a curved path with constant horizontal velocity and changing vertical velocity due to gravity. Factors like initial velocity and launch angle affect the motion. An example problem calculates the horizontal distance traveled and height of an object launched at an angle after 2 seconds.

1. What is Projectile
Motion?

2. Instructional Objectives:
• Students will be able to:
– Define Projectile Motion
– Distinguish between the different types of
projectile motion
– Apply the concept to a toy car and measure its
velocity

3. Projectile Motion
• Two-dimensional motion of an object
– Vertical
– Horizontal

4. Types of Projectile
Motion
• Horizontal
– Motion of a ball rolling freely along a
level surface
– Horizontal velocity is ALWAYS
constant
• Vertical
– Motion of a freely falling object
– Force due to gravity
– Vertical component of velocity
changes with time
• Parabolic
– Path traced by an object accelerating
only in the vertical direction while
moving at constant horizontal velocity

5. Examples of Projectile
Motion
• Launching a Cannon ball

7. Equations
• X- Component
• Y- Component
• Vectors
tvxx xiif 
gtvv
ygvv
gttvyy
yiyf
yiyf
yiif



2
2
1
22
2
)sin(
)cos(


iyi
ixi
vv
vv


Note: g= 9.8
m/s^2

8. Factors Affecting
Projectile Motion
• What two factors would affect projectile
motion?
– Angle
– Initial velocity
Initial Velocity
Angle

9. Class Exercise
An object is fired from the ground at 100
meters per second at an angle of 30
degrees with the horizontal
Calculate the horizontal and vertical
components of the initial velocity
After 2.0 seconds, how far has the object
traveled in the horizontal direction?
How high is the object at this point?

10. Solution
• Part a
• Part b
• Part c
   ms
s
mtvx
t
x
v
x
ix
1740.287 



  
   s
m
s
mvv
s
m
s
mvv
iiy
iix
5030sin100sin
8730cos100cos
0
0




       2
2
2
0.28.9
2
1
0.250
2
1
s
s
ms
s
mtgtvy iy 

11. Applications
Any Ideas?

12. THANK YOU