A tennis ball is released from a height of hcm, it rebounds one-third of the distance it has fallen after each fall.

Find the total distance travelled when it strikes the ground
(i) the third time
(ii) the nth time

1 answer

for our purposes, consider h=1. The round trip on each bounce is 2 times the height achieved. So, the total distance traveled is

1 + 2/3 + 2/9 + 2/27 + ...
with sums adding up to
1, 5/3, 17/9, 53/27

The round trip distance after the nth bounce is thus 2/3^n

The total distance traveled is thus

3(1-1/(3^n+1)) - 1 = (2*3^n - 1)/3^n

Check:
n=0: 1
n=1: 5/3
n=2: 17/9
n=3: 53/27