A highly elastic ball is released from rest at a distance h above the ground and bounces up and down. With each bounce a fraction$\text{f}\text{=}\text{1/10}$  of its KE just before the bounce is lost. The time for which the ball will continue to bounce is nearly$\left(\text{take}\text{g}\text{=10}{\text{m/s}}^2,\text{h}=5\text{m}\right)$

${\text{h}}_{\text{0}}=5\text{m}$

$\text{u}=\sqrt{2{\text{gh}}_{\text{0}}}=\sqrt{2\times 10\times 5}$

$=10\text{m/s}$

${\text{v}}_1=\text{eu}$

Fraction of loss in K.E $\text{=}\frac{1}{10}\left(\text{given}\right)$

$\frac{\frac{1}{2}{\text{mu}}^{\text{2}}-\frac{1}{2}{\text{mv}}_{\text{1}}^{\text{2}}}{\frac{1}{2}{\text{mu}}^{\text{2}}}=\frac{1}{10}$

$\frac{{\text{u}}^{\text{2}}-{\text{e}}^{\text{2}}{\text{u}}^{\text{2}}}{{\text{u}}^{\text{2}}}=\frac{1}{10}$

$1-{\text{e}}^{\text{2}}=\frac{1}{10}$

${\text{e}}^2=\frac{9}{10}$

$\text{e}=\frac{3}{\sqrt{10}}$

$\text{e}=\frac{3}{3.1622}$

$=0.9488$

$\simeq 0.949$

Substitute  in (1)                      

$\text{T}=\sqrt{\frac{2\times 5}{10}}\left(\frac{1+0.949}{1-0.949}\right)$

$=\left(\frac{1.949}{0.051}\right)$

$=38.215$

$\text{T}=38\sec$