1. Projectile Motion YouTube - Baxter NOOOOOOOOOO

2. Projectile Motion 9.1Projectile motion (AHL) 9.1.1State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. 9.1.2Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance. 9.1.3Describe qualitatively the effect of air resistance on the trajectory of a projectile. 9.1.4Solve problems on projectile motion.

3. 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.

4. Amazing facts!

8. Why?

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

11. Vertical and horizontal independent of each other 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. Horizontally velocity remains constant

12. Projectiles (Half Trajectory) An object projected sideways through the air will follow a curved trajectory. horizontal motion (steady speed) vertical motion The horizontal and vertical motions should be treated separately. Time is the only quantity common to both. accelerates downwards at -9.8 ms -2

13. At any point in its trajectory, the velocity of a projectile has two components. one vertical, V V the other horizontal, V H The resultant velocity is found drawing a vector diagram and add the vectors together, TIP to TAIL. Vector Diagram horizontal velocity vertical velocity resultant/actual velocity

14. vhvh v This is an example of a ‘half-trajectory.’ GREEN – actual motion RED – vertical motion BLUE – horizontal motion

15. Watch that dog! Imagine a dog being kicked horizontally off the top of a cliff (with an initial velocity v h ). vhvh

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

17. Parabola

18. Why?

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

20. Horizontal motion Assuming no air resistance, there are no horizontal forces. This means horizontally the dog moves with constant speed v h vhvh Horizontal distance travelled (x) = v h t

21. Vertical motion Assuming no air resistance, there is constant force downwards (=mg). This means vertically the dog moves with constant acceleration g = 9.8 m.s -2 Vertical distance travelled (y) = u v t + ½gt 2

22. Parabolic motion Since y = ½gt 2 (if u = 0) and x = v h t, y = ½gx 2 /v h 2 which you may (!) recognise as the formula of a parabola. Another piece of ultra cool physics!

23. 30 ms -1 Example A ball is kicked horizontally off an embankment, with a velocity of 30 ms -1. It lands 24 m from the base of the embankment. (a)Calculate how long the ball was in flight. 24 m common to horizontal and vertical motions

24. HorizontalVertical (b)Calculate the horizontal velocity just before hitting the ground. travels horizontally at steady speed – no acceleration horizontally not initially falling down, so speed of zero in vertical direction acted upon by gravity Horizontal

25. (c)Calculate the vertical velocity just before hitting the ground. Vertical (d)How high is the embankment? Vertical means 7.84 ms -1 downwards so height of the embankment is 3.14 m means ball fell through distance of 3.14 m

26. (e)Calculate the resultant velocity of the ball, just before hitting the ground. 30 ms -1 -7.8 ms -1 velocity Size By Pythagoras:

27. Direction 30 ms -1 -7.8 ms -1 velocity

28. Q1.A ball is kicked off a cliff with a horizontal speed of 16 ms -1. The ball hits the ground 2.2 s later. (a)Calculate the height of the cliff. (b)Calculate the distance between the foot of the cliff and where the ball lands. (c)Calculate the vertical component of the balls velocity just before it hits the ground. (d)Calculate the balls velocity as it hits the ground. 23.7 m 35.2 m 21.6 ms -1 26.9 ms -1 at angle of 53.5° below horizon You may want to draw a diagram to help you get started !!!

29. Q2.A ball is kicked off a cliff with a horizontal speed of 22 ms -1. the ball hits the ground 1.5 s later. (a)Calculate the height of the cliff. (b)Calculate the horizontal distance from the foot of the cliff, to where the ball lands. (c)Calculate the vertical component of the balls velocity as it hits the ground. (d)Calculate the balls actual velocity as it hits the ground. 11 m 14.7 ms -1 26.5 ms -1 at angle of 34° below horizon 33 m You may want to draw a diagram to help you get started !!!

30. Example A dog 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 dog hit the ground? 5 m.s -1 30 m

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

32. 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 dog hits the ground 12.4 metres from the base of the cliff 5 m.s -1 30 m

33. Parabola 12.4 metres

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

35. What is the dog’s speed as he hits the ground? Total energy at top = ½mv 2 + mgh Total energy = ½m(5) 2 + m x 9.8 x 30 Total energy = 12.5m + 294m = 306.5m 5 m.s -1 30 m

36. What is the dog’s speed as he hits the ground? At the bottom, all the potential energy has been converted to kinetic energy. All the dog’s energy is now kinetic. V = ? energy = ½mv 2

37. What is the dog’s speed as he hits the ground? energy at top = energy at bottom 306.5m = ½mv 2 306.5 = ½v 2 613 = v 2 V = 24.8 m.s -1 (Note that this is the dog’s speed as it hits the ground, not its velocity. v = 24.8 m.s -1

38. Projectiles (Full Trajectory) A projectile does not need to be an object falling, but could be an object fired at angle to the horizontal. The subsequent motion would be max height θ

39. If air resistance is ignored, the trajectory has an axis of symmetry about the mid point (maximum height). So the time taken to reach the maximum height is the same as the time taken to fall back to the ground. Various calculations can be made, but firstly, the initial velocity must be split into its horizontal and vertical components. Horizontal a = 0 ms -2 Vertical a = -9.8 ms -2

40. Starting with non-horizontal motion Woof! (help)

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

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

43. Starting with non-horizontal motion 2. Looking at the vertical motion, when the dog hits the floor, displacement = 0 Initial vertical velocity = v v = 25sin30° Acceleration = - 9.8 m.s -2 30° 25 m.s -1

44. Starting with non-horizontal motion 3. Using s = ut + ½at 2 0 = 25sin30°t + ½(-9.8)t 2 0 = 12.5t - 4.75t 2 0 = 12.5 – 4.75t 4.75t = 12.5 t = 12.5/4.75 = 2.63 s 30° 25 m.s -1

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

46. 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 2 30°

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

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

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

50. Starting with non-horizontal motion 6. So ½m(25cos30°) 2 + mgh = ½m(25) 2 ½(21.65) 2 + 9.8h = ½(25) 2 234.4 + 9.8h = 312.5 9.8h = 78.1 h = 8.0 m 30°