A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back 60% of the height from which it fell. It continues bouncing in this way− each time rising to 60% of the previous height.
(i) What height does the ball reach after the \(5^{th}\) bounce?
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the \(6^{th}\) time?

Initial height = 80 m
Common ratio, r = 60%

The bounce heights form a GP : \(48, 28.8, 17.28, 10.368, 6.2208,...\)
where
\(a_1 = 80(0.6) = 48 m\)

Using the \(n^{th}\) term formula
\(a_n = a_1 r^{n−1}\)
For the \(5^{th}\) bounce,
\(a_5 = 48(0.6)^4 = 48(0.296) = 6.2208 m\)
After the \(5^{th}\) bounce, the ball reached a height of 6.2208 metres.

The ball:
(a) falls 80 m initially
(b) then each bounce consists of travelling up and down
By the time the ball hits the ground for the \(6^{th}\) time, it has completed 5 bounces.
So, total distance = 80 + 2(sum of first 5 bounce heights)
To find the sum of first 5 bounces, use the sum formula
\(S_n = \frac{a_1(1 − r^n)}{1 − r}\)
where
\(a_1 = 48,\,r = 0.6,\,n = 5\)
\(S_5 = \frac{48(1 − 0.6^5)}{1 − 0.6}\, = \, 110.6688 m\)
Therefore, total distance \(= 80 + 2(110.6688) = 301.3376 m\)