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Q. After striking the floor a certain ball rebounds $\frac{4}{5} \text{th}$ of its height from which it has fallen. The total distance that the ball travels before coming to rest if it is gently released from a height of $120 \,m$ is

BITSATBITSAT 2017

Solution:

Initially, the ball falls from a height of $120 \,m$.
After striking the floor, it rebounds and goes to a height of $\frac{4}{5} \times(120) \,m$.
Now, it falls from a height of $\frac{4}{5} \times(120) m$ and after rebounding goes to a height of $\frac{4}{5}\left(\frac{4}{5}(120)\right)\, m$.
This process is continued till the ball comes to rest.
Hence, the total distance travelled is
$D =120+2\left[\frac{4}{5}(120)+\left(\frac{4}{5}\right)^{2}(120)+\ldots \infty\right]$
The above series in brackets is an infinte $GP$,
$\therefore D =120+2\left[\frac{\frac{4}{5}(120)}{1-\frac{4}{5}}\right]$
$=120+2\left[\frac{480}{1}\right]=1080\, m$