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 \(y\).
Show that
$$y = x \tan \alpha - \frac { g x ^ { 2 } } { 2 u ^ { 2 } \cos ^ { 2 } \alpha }$$
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 \(7 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at an angle of elevation of \(45 ^ { \circ }\). By modelling the ball as a particle moving freely under gravity,
find the horizontal distance of the ball from \(A\) when the ball is 1 m above the beach.
A boy is standing on the beach at the point \(B\) vertically below \(A\). He starts to run in a straight line with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), leaving \(B 0.4\) seconds after the ball is thrown.
He catches the ball when it is 1 m above the beach.