1. Projectile motion

2. What is projectile motion? Projectiles are objects that are thrownor catapulted They move vertically and horizontally, following a parabolic path Think kicking a ball in the air or something rolling over a cliff

3. Forces on projectiles Projectiles have only one force acting on them – gravity, which always acts downwards This means that they move sideways at a constant speed, but they accelerate up or down

4. Sample question A marble rolls off the edge of a table with constant speed. Describe the path it will follow and explain why it will follow this path

5. Sample answer The marble will follow a parabolic path The marble has constant horizontal velocity. There are no forces acting horizontally, so it will not accelerate in this direction The marble accelerates downwards. Gravitational force acts vertically, so it will accelerate in this direction Because it has no vertical velocity to start with, but accelerates downwards, it will follow a parabolic path

6. Check our answer We said what the shape will be We said why it has constant horizontal speed We said why it has vertical acceleration We linked the two parts of the motion to the shape

7. Solving problems with projectiles To solve problems with projectiles, we divide the motion into two parts – the vertical part and the horizontal part The key to this is remembering that the time for each part must be the same

8. Splitting projectile motion into parts It seems odd that the time is the same, but consider two objects dropped from the same height – do they reach the ground at the same time? Now think about two objects, one rolled off the edge of a table and one dropped – do they travel the same vertical distance? Yes, so they must reach the ground at the same time Try this yourself at home, using marbles or tennis balls.

9. Finding the initial velocities We have to use trigonometry to split the motion of an object into its two parts The initial velocity is the hypotenuse of the triangle, as in the diagram below: Initial velocity Vertical component of the velocity HypotenuseOpposite Angle of launch Horizontal component of the velocity Adjacent

10. Sample question A ball is kicked with an initial velocity of 13ms -1 at an angle of 30 o relative to the ground, as shown below. Find the horizontal and vertical components of the velocity 13ms -1 30 o

11. Sample answer Vertical component: sinθ = opposite/hypotenuse sin30 = opp/13 opp = 13xsin30 opp = 5.90ms -1 Horizontal component: cosθ=adjacent/hypotenuse cos30 = adj/13 adj = 13xcos30 adj = 11.6ms -1

12. Check our answer Looking at the diagram, we expect the vertical component to be less than the horizontal Neither component is bigger than the original number Neither component is really small compared to the original Our answers have the correct units

13. Types of projectile problems Most projectile problems are solved the same way You might be asked to find: – The range (total distance covered) or where an object lands – The maximum height reached – Whether something goes over or under an object a certain distance away

14. Solving projectile problems – vertical part If needed, find the initial vertical and horizontal velocities (not required for objects rolling/being thrown from a height) Vertical motion has acceleration, so use kinematic equations to find: – The time taken to the peak height (where the velocity is zero for a moment) – The peak height (if needed)

15. Solving projectile problems – horizontal part There is no horizontal acceleration, so once you have the time, you can find the range using v=d/t Make sure the velocity in this step is the horizontal velocity!

16. Sample question A ball is rolled off a table of height 1.3m with an initial horizontal velocity of 1.35ms-1. A target is placed 0.55m from the table. Will the ball land on the target?

17. Sample answer Time taken to hit the ground: d=1.3md=v i t+½at 2 v i =01.3=0+0.5x9.8t 2 a=9.8ms-2 t=√1.3/4.9 t=0.515s

18. Sample answer Horizontal distance travelled in that time v=1.35ms -1 v=d/t t=0.515s2.35=d/0.515 d=2.35x0.515 d=0.69m The ball will not land on the target

19. Check answer The time found is about right for dropping something off a table, based on experience The time taken is positive (remember, sometimes acceleration due to gravity is negative depending on whether the object is falling or rising compared to the origin) The final answer is reasonable The answer has correct units The question asked has been answered

20. Sample question A plane with horizontal velocity of 110ms -1 drops a parcel to people stranded on an island 45m below. From how far away should they drop the parcel?

21. Sample answer Time taken for parcel to fall to land d=45md=v i t+½at 2 v i =045=0+0.5x9.8t 2 a=9.8ms -2 t=√45/4.9 t=3.03s

22. Sample answer Horizontal distance travelled in that time: v=110ms-1v=d/t t=3.03s110=d/3.03 d=110x3.03 d=333m

23. Check the answer It’s harder to check answers in situations which are outside our everyday experience Check that your rearranging is correct, particularly the square root in the kinematic equation (and that you did the multiplication before you square rooted the answer) Your units must be correct

24. Sample question Hemi stands 20m from a goal with a crossbar 5.1m high. If he kicks the ball with an initial velocity of 18.5ms -1 at an angle of 40 o, find if the ball will go over the crossbar.

25. Sample answer Find the initial horizontal and vertical velocities Vertical:sinθ=opp/hyp sin40=opp/18.5 opp=18.5xsin40 opp=10.9ms -1 Horizontal:cosθ=adj/hyp cos40=adj/18.5 adj=18.5xcos40 adj=15.0ms -1

26. Sample answer Work out the time taken to reach the crossbar d=20mv=d/t v=15.0ms -1 15=20/t t=20/15 t=1.33s

27. Sample answer Time taken to reach maximum height v i =10.9ms -1 v f =v i +at v f =00=10.9+9.8t a=9.8ms -2 t=-10.9/-9.8 t=1.11s Time difference=0.22s from the maximum height

28. Sample answer Maximum height v i =10.9ms -1 d= ½ (vf+vi)t t=1.11sd=0.5x10.9x1.11 v f =0d=6.05m Distance dropped in 0.22s vi=0d=v i t+½at 2 t=0.22sd=0+0.5x9.8x0.22 2 d=1.08m

29. Sample answer Hence, height when the ball reaches the crossbar Height = 6.05 – 1.08 = 4.97m The ball will not go over the crossbar

30. Check the answer There were a lot of steps in this solution, so check that you’ve transferred all the numbers from one section to the next correctly Check that the answers have the correct units