M2 June 2011 Q8
8. A particle is projected from a point \(O\) with speed \(u\) at an angle of elevation \(\alpha\) above the horizontal and moves freely under gravity. When the particle has moved a horizontal distance \(x\), its height above \(O\) is \(y\).
A girl throws a ball from a point \(A\) at the top of a cliff. The point \(A\) is 8 m above a horizontal beach. The ball is projected with speed 7 m s\(^{-1}\) at an angle of elevation of 45\(^\circ\). By modelling the ball as a particle moving freely under gravity,
A boy is standing on the beach at the point \(B\) vertically below \(A\). He starts to run in a straight line with speed \(v\) m s\(^{-1}\), leaving \(B\) 0.4 seconds after the ball is thrown.
He catches the ball when it is 1 m above the beach.
| Scheme | Marks |
|---|---|
| Horiz: \(\ x = u\cos\alpha t\) | B1 |
| Vert: \(\ y = u\sin\alpha t - \dfrac{1}{2}gt^2\) | M1 |
| \(y = u\sin\alpha \times \dfrac{x}{u\cos\alpha} - \dfrac{1}{2}g \times \dfrac{x^2}{u^2\cos^2\alpha}\) | DM1 |
| \(y = x\tan\alpha - \dfrac{gx^2}{2u^2\cos^2\alpha}\quad **\) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(y = -7:\quad -7 = \tan 45x - \dfrac{gx^2}{2 \times 7^2\cos^2 45}\) | M1 A1 |
| \(-7 = x - \dfrac{9.8x^2}{7^2}\) | |
| \(-7 = x - \dfrac{x^2}{5}\) | M1 |
| \(x^2 - 5x - 35 = 0\) | |
| \(x = \dfrac{5 \pm \sqrt{25 + 4 \times 35}}{2}\) | M1 |
| \(x = 8.92\) or 8.9 | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Time to travel 8.922 m horizontally \(= \dfrac{8.922}{7\cos 45}\ = 1.802\ldots\) s | M1 |
| \(v = \dfrac{8.922}{1.402}\) | M1 A1 ft |
| \(= 6.36\) or \(6.4\ \left(\text{m s}^{-1}\right)\) | A1 |
| (4) | |
| (13 marks) |