M1 June 2010 Q6
6. A ball is projected vertically upwards with a speed of 14.7 m s\(^{-1}\) from a point which is 49 m above horizontal ground. Modelling the ball as a particle moving freely under gravity, find
(a) the greatest height, above the ground, reached by the ball, (4)
(b) the speed with which the ball first strikes the ground, (3)
(c) the total time from when the ball is projected to when it first strikes the ground. (3)
| Scheme | Marks |
|---|---|
| \((\uparrow)\ \ v^2 = u^2 + 2as\) | |
| \(0 = 14.7^2 - 2 \times 9.8 \times s\) | M1A1 |
| \(s = 11.025\) (or 11 or 11.0 or 11.03) m | A1 |
| Height is 60 m or 60.0 m ft | A1ft |
| (4) |
| Scheme | Marks |
|---|---|
| \((\downarrow)\ \ v^2 = u^2 + 2as\) | |
| \(v^2 = (-14.7)^2 + 2 \times 9.8 \times 49\) | M1 A1 |
| \(v = 34.3\) or 34 m s\(^{-1}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \((\downarrow)\ \ v = u + at\) OR \((\downarrow)\ \ s = ut + \tfrac{1}{2}at^2\) | |
| \(34.3 = -14.7 + 9.8t\) \(49 = -14.7t + 4.9t^2\) | M1 A1 |
| \(t = 5\) \(t = 5\) | A1 |
| (3) | |
| (10 marks) |