1. Physics 12
Projectile Motion

2. Clip of the day: minuteEarth instead today!

3. Projectile Motion
We know that an object (in the absence of air resistance) that is launched at a given angle should follow a parabolic path
We need to break the velocities into x and y components

4. Review: Projectile Motion – Horizontal Launch
An object that is launched horizontally will have no initial velocity in the y direction so the entire initial velocity will be in the x direction
At this point, we are able to treat the projectile using our two equations of motion

5. Equations of Motion
However, in 2D, we need four EOM’s to separate the horizontal and vertical components of motion:
Recall in 1D:

6. Projectile Motion – Simplified EOMs
Considering there is no acceleration in x and the acceleration in y is g (-9.81m/s2)

7. Review: Example 1
A cannonball is fired horizontally from the top of a 50.m high cliff with an initial speed of 30.m/s. Ignoring air resistance, determine the following:
How long it takes to strike the ground
How far from the base of the cliff it strikes the ground
How fast it is travelling when it strikes the ground

8. Start with y position equation (4)
Sub in known information (h=50.m) and solve for time

9. Now use x position equation (2)
Sub in time and known information (t=3.2s, vox=30.m/s) and solve for dx

10. Finally we will use equations (1) and (3)
Sub in time and solve for velocity

11. Now, we employ trigonometry and Pythagorean Theorem to solve for the final velocityvx vy vr

12. Non-horizontal launch:
Watch video and notice what happens to the velocities
Notice: Vx does not change and Vy goes from positive to zero at vertex to negative

13. Example 2:
A golfer uses a club that launches a golf ball at a 15° angle at a speed of 45m/s. Determine the following:
The time the golf ball is in the air
The horizontal distance the ball travels
The velocity as it strikes the ground
The maximum height the ball attains

14. Start with the position in the y equation
Include the initial velocity in the y term
Solve for when the position in the y is equal to zero

15. Use the time from the previous question and the position in the x equation
Solve for the range (horizontal distance)

16. Solve for the velocity in the x and y direction

17. Use Pythagorean Theorem and Trig to solve for final velocity

18. Max height will occur when y velocity is equal to zero
Max height will occur when y velocity is equal to zero. Solve for time and then sub into y position equation

19. A few more to try:
A boy kicked a can horizontally from a 6.5 m high rock with a speed of 4.0 m/s. How far from the base of the rock the can land?
A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
A cannon that launches a cannon ball at a 26° angle at a speed of 33m/s. Determine the following:
The time the cannon ball is in the air
The horizontal distance the cannon ball travels
The velocity as it strikes the ground
The maximum height the cannon ball attains
Page 536
Questions 1-8 ANSWERS: 1) 5.9 m 2) 16.5 m/s
3) a) 2.85sec b)86m c)33m/s,334° d) 10m