2019 NYJC P2 Q3

When a ball is dropped from a height of \(H\)m above the ground, it will rebound to a height of \(eH\)m where \(0 < e < 1\). The height of each successive bounce will be \(e\) times of that of its previous height. It is also known that the time taken between successive bounce is given by \(t=0.90305\sqrt{h}\) where \(h\) is the maximum height of the ball from the ground between these bounces. We can assume that there is negligible air resistance.

A ball is now dropped from a height of \(10\)m from the ground. Let \({{t}_{n}}\) be the time between the \({{n}^{\mathrm{th}}}\) and \({{\left( n+1 \right)}^{th}}\) bounce.

1. Show that the total distance travelled by the ball just before the \({{n}^{\mathrm{th}}}\) bounce is \(\frac{10\left( 1+e+2{{e}^{n}} \right)}{1-e}\).[3]
2. Show that \({{t}_{n}}\) is a geometric sequence. State the common ratio for this sequence.[3]
3. Find in terms of \(e\) the total distance the ball will travel and the time taken when it comes to rest. You may assume that between any two bounces, the time taken for the ball to reach its maximum height is the same as the time it takes to return to the ground.[3]