1. Projectile Motion
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2. Amazing facts!
If a gun is fired horizontally, and at the same time a bullet is dropped from the same height. They both hit the ground at the same time.

3. Amazing facts!

4. Amazing facts!

5. Amazing facts!

6. Amazing facts!
Why?

7. Vertical and horizontal
Their vertical motion can be considered separate from their horizontal motion.

8. Vertical and horizontal
Vertically, they both have zero initial velocity and accelerate downwards at 9.8 m.s-2. The time to fall the same vertical distance is therefore the same.

9. Watch that ball!
Imagine a ball being kicked horizontally off the top of a cliff (with an initial velocity vh). vh

10. Parabola
Assuming that there is negligible air resistance, he falls in the path of a parabola.

11. Parabola

12. Parabola
Why?

13. Why a parabola?
We can consider the motion to be the sum of his horizontal motion and vertical motion.
We can treat these separately vh

14. Horizontal motion
Assuming no air resistance, there are no horizontal forces.
This means horizontally
the ball moves with
constant speed vh vh
Horizontal distance travelled (x) = vht

15. Vertical motion
Assuming no air resistance, there is constant force downwards (=mg).
This means vertically the
ball moves with constant
acceleration g = 9.8 m.s-2
Vertical distance travelled (y) = uvt + ½gt2

16. Parabolic motion Since y = ½gt2 (if u = 0) and x = vht,
y = ½gx2/vh2 which you may (!) recognise as the formula of a parabola.

17. Example
A ball is kicked off the top of a cliff with an initial horizontal velocity of 5 m.s-1. If the cliff is 30 m high, how far from the cliff bottom will the ball hit the ground?
5 m.s-1
30 m

18. Example Looking at vertical motion first: s = ut + ½at2
u = 0, a = 9.8 m.s-2, s = 30 m, t = ?
s = ut + ½at2
30 = ½ x 9.8 x t2
t2 = 6.1
t = 2.47 s
The ball hits the ground after 2.47 seconds (yes!)
5 m.s-1
30 m

19. Example Now look at horizontal motion:
Constant speed (horizontally) = 5 m.s-1
Time of fall = 2.47 seconds
Horizontal distance travelled = speed x time
Horizontal distance travelled = 5 x 2.47
= 12.4 m The ball hits the ground 12.4 metres from the base of the cliff
5 m.s-1
30 m

20. Parabola
12.4 metres

21. What is the ball’s speed as it hits the ground?
To answer this it is easier to think in terms of the ball’s total energy (kinetic and potential)
5 m.s-1
30 m

22. What is the ball’s speed as it hits the ground?
Total energy at top = ½mv2 + mgh
Total energy = ½m(5)2 + mx9.8x30
Total energy = 12.5m + 294m = 306.5m
5 m.s-1
30 m

23. What is the ball’s speed as it hits the ground?
At the bottom, all the potential energy has been converted to kinetic energy. All the ball’s energy is now kinetic.
energy = ½mv2
V = ?

24. What is the ball’s speed as it hits the ground?
energy at top = energy at bottom
306.5m = ½mv2
306.5 = ½v2
613 = v2
V = 24.8 m.s-1
(Note that this is the ball’s
speed as it hits the ground,
not its velocity.
v = 24.8 m.s-1

25. Starting with non-horizontal motion

26. Starting with non-horizontal motion
25 m.s-1
30°

27. Starting with non-horizontal motion
Split the initial velocity into vertical and horizontal components
vh = 25cos30°
vv = 25sin30°
25 m.s-1
30°

28. Starting with non-horizontal motion
2. Looking at the vertical motion, when the ball hits the floor, displacement = 0
Initial vertical velocity = vv = 25sin30°
Acceleration = m.s-2
25 m.s-1
30°

29. Starting with non-horizontal motion
3. Using s = ut + ½at2
0 = 25sin30°t + ½(-9.8)t2
0 = 12.5t t2
0 = 12.5 – 4.75t
4.75t = 12.5
t = 12.5/4.75 = 2.63 s
25 m.s-1
30°

30. Starting with non-horizontal motion
4. Looking at horizontal motion
Ball in flight for t = 2.63 s travelling with constant horizontal speed of
vh = 25cos30° = 21.7 m.s-1.
Distance travelled = vht = 21.7x2.63 = 57.1m
30°
57.1m

31. Starting with non-horizontal motion
5. Finding maximum height? Vertically;
v = 0, u = 25sin30°, t = 2.63/2
s = (u + v)t = 12.5x1.315 = 8.2m
30°

32. Starting with non-horizontal motion
6. Don’t forget some problems can also be answered using energy.
30°

33. Starting with non-horizontal motion
6. Don’t forget some problems can also be answered using energy.
As ball is fired total energy = ½m(25)2
25 m.s-1
30°

34. Starting with non-horizontal motion
6. At the highest point,
total energy = KE + GPE =½m(25cos30°)2 + mgh
As ball is fired total energy = ½m(25)2
30°

35. Starting with non-horizontal motion
6. So ½m(25cos30°)2 + mgh = ½m(25)2
½(21.65) h = ½(25)2
h = 312.5
9.8h = 78.1
h = 8.0 m
30°

36. Investigation