Question

A ball is thrown vertically upward from ground with speed $40m/s$. It collides with ground after returning. Find the total distance traveled and time taken during its bouncing. ($e=0.5$) ($g=10m/s^2$)

Answer 1

Distance traveled before first collision $=\frac{u^2\times 2}{2g}=160m$

Time taken $=\sqrt{\frac{2\times 160}{10}}=\sqrt{32}sec$

After collision velocity will reduce to half because velocity of separation=velocity of approach

Distance traveled before third collision $=(\frac{u}{2}{)}^2\times \frac{2}{2g}=40m$

Time taken $=\frac{\sqrt{3}2}{\sqrt{2}}$

Total distance after infinite collision $=160(1+\frac{1}{4}+\frac{1}{16}......\infty )$

$=160\left(\frac{4}{3}\right)m=213.3m$

Total time taken$=\sqrt{32}[1+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}....\infty ]$

$=\sqrt{32}\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)$

$=12.4sec$