M1 June 2011 Q1
1. At time \(t = 0\) a ball is projected vertically upwards from a point \(O\) and rises to a maximum height of 40 m above \(O\). The ball is modelled as a particle moving freely under gravity.
(a) Show that the speed of projection is 28 m s\(^{-1}\). (3)
(b) Find the times, in seconds, when the ball is 33.6 m above \(O\). (5)
| Scheme | Marks |
|---|---|
| \(0^2 = u^2 - 2 \times 9.8 \times 40\) | M1 A1 |
| \(u = 28\) m s\(^{-1}\) ** GIVEN ANSWER | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(33.6 = 28t - \tfrac{1}{2}9.8t^2\) | M1 A1 |
| \(4.9t^2 - 28t + 33.6 = 0\) | |
| \(t = \dfrac{28 \pm \sqrt{28^2 - 4 \times 4.9 \times 33.6}}{9.8}\) | M1 |
| \(= 4\) s or (1.7 s or 1.71 s) | A1 A1 |
| (5) | |
| (8 marks) |