1. Projectiles

2. 50 40 o A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course.

3. 50 40 o Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2. What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2

4. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

5. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

6. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

7. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

8. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

9. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

10. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

11. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

12. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

13. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

14. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

15. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

16. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o What are the horizontal and vertical components of acceleration? The horizontal component is zeroThe vertical component is –10. s = ut + ½at 2 (1) (2) v = u + at (3) (4) Write down the vector equations for the position and velocity at time t seconds after the ball is hit. Take g to be -10ms -2.

17. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o Find the greatest height reached y = H At max. height H, v y = 0Use equation (4) (1) (2) (3) (4)

18. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o Find the greatest height reached y = H At max. height H, v y = 0 Find the greatest height reached (1) (2) (3) (4)

19. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o y = H At max. height H, v y = 0 Find the greatest height reached (1) (2) (3) (4)

20. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o y = H At max. height H, v y = 0 Find the greatest height reached (1) (2) (3) (4)

21. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o y = H At max. height H, v y = 0 Now use equation (2) Find the greatest height reached (1) (2) (3) (4)

22. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o Find time taken to reach the ground y = H In this particular case, using the symmetrical properties of the parabola, t = 2  3.21 = 6.42 secs (1) (2) (3) (4)

23. A golf ball is hit with a speed of 50ms -1 at an elevation of 40 o along a level course. 50 40 o Find the horizontal distance from the start to the landing point y = H 6.42 secs R = x The horizontal distance, or range is R = x. Use (1) (1) (2) (3) (4)

24. A ball is thrown from the top of a cliff 52m above sea level. With the x axis taken horizontally, the y axis taken vertically upwards and the origin at the point of projection, the velocity of projection is given by Find the speed and angle of elevation of the ball at the start x y (1) (2) (3) (4) Use (3) and (4) with t = 0 Speed Speed = 25ms -1 If the angle of elevation is , then

25. A ball is thrown from the top of a cliff 52m above sea level. With the x axis taken horizontally, the y axis taken vertically upwards and the origin at the point of projection, the velocity of projection is given by Find the time of the flight x y (1) (2) (3) (4) Find t when y = -52 by substituting in (2)  t = 4 seconds

26. A ball is thrown from the top of a cliff 52m above sea level. With the x axis taken horizontally, the y axis taken vertically upwards and the origin at the point of projection, the velocity of projection is given by Find the distance from the foot of the cliff to the point where the ball lands x y (1) (2) (3) (4) We have already found that the duration of the fight was 4 seconds, so substitute t = 4 into (1) x = 24(4) = 96 m

27. A ball is thrown from the top of a cliff 52m above sea level. With the x axis taken horizontally, the y axis taken vertically upwards and the origin at the point of projection, the velocity of projection is given by Find the velocity when the ball lands and hence the speed and angle of entry x y (1) (2) (3) (4) Substitute t = 4 into (3) and (4) Speed Speed = 40.8 ms -1 If the angle of elevation is , then