1. Projectile Motion

2. Examples of Projectile Motion:
A cannon ball
Throwing a baseball
A ball rolling off a table
Throwing a rock
The list can go on and on.

3. What is a projectile?
A projectile is an object upon which the only force acting on it is gravity.
A projectile is any object that once projected continues in motion by its own inertia and is influenced ONLY by gravity.

5. Common misconceptions
Many students ask “How can an object be moving upward if the only force acting on it is gravity?”
Trust Newton
A force is NOT necessary to keep an object in motion
A force is only necessary to keep an object in ACCELERATION

8. A projectile has two vector components: horizontal and vertical (x and y)
These two components of motion must be discussed separately

9. Acceleration due to gravity (g) only affects the vertical motion of the projectile.
The horizontal motion is unaffected by gravity.

10. To Summarize

11. Non- Horizontally Launched
If a projectile is launched non-horizontally:

12. With gravity, the object falls with a parabolic trajectory or path
The projectile still moves the same horizontal distance in each second travelled as it did when gravity was NOT a factor
Gravity only affects the vertical

14. Check your understanding

15. Supposing a snowmobile is equipped with a flare launcher that is capable of launching a flare vertically (relative to the snowmobile). If the snowmobile is in motion and launches the flare and maintains a constant horizontal velocity after the launch, then where will the flare land (neglect air resistance)?
a. in front of the snowmobile
b. behind the snowmobile
c. in the snowmobile

16. http://www.physicsclassroom.com/mmedia/vectors/tb.gi f

17. Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane?
a. below the plane and behind it.
b. directly below the plane
c. below the plane and ahead of it

18. http://www.physicsclassroom.com/mmedia/vectors/pap. gif

19. Numbers
So we’ve been working with displacement a lot in terms of vector addition. But velocity is a vector, too
A boat moves with velocity 4 m/s across a river
The current is moving downstream. What does the boat’s velocity relative to an observer on the shore look like?

20. Review:
A projectile is any object upon which the only force is gravity,
Projectiles travel with a parabolic trajectory due to the influence of gravity,
There are no horizontal forces acting upon projectiles and thus no horizontal acceleration,
The horizontal velocity of a projectile is constant (never changing in value),
There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down,
The vertical velocity of a projectile changes by 9.8 m/s each second,
The horizontal motion of a projectile is independent of its vertical motion.

21. Consider the cannon on the cliff again:
Assume it fires the cannonball at 20 m/s
Gravity causes the cannonball to accelerate downwards at 9.8 m/s/s

23. Important Concept: Horizontal velocity stays the same
Vertical velocity changes by 9.8 m/s every second

24. http://www.physicsclassroom.com/mmedia/vectors/hlp. gif

25. How about an upward angle?
75.7 m/s at an angle of 15o

26. http://www.physicsclassroom.com/mmedia/vectors/nhlp .gif

27. How to find initial velocity?
Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal.

28. Horizontal and Vertical Displacement
Vertical displacement (height) can be calculated using
y = viyt +½ gt2
The horizontal displacement (distance or range) can be calculated with
x = vixt

29. Example:

30. Time
tP = viy/g ttotal = 2 tP

31. In summary

33. Sample Problem
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.
Horizontal Info: x= 35 m, vix=?, ax= 0 m/s2
Vertical Info: y =-22 m, viy=0 m/s, ay= -9.8 m/s2

34. Use y = viy • t + 0.5 • ay • t2 to solve for time; the time of flight is 2.12 seconds.
Now use x = vix • t • ax • t2 to solve for vix
Note that ax is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12 s for t, the vix can be found to be m/s.