A body is dropped from a height of 16m. The body strikes the ground and losses $25\%$ of its velocity. The body rebounds to a height of

When the body is dropped, then u = 0, distance travelled is 16 m therefore final velocity before hitting the ground is

${\text{v}}^2={\text{u}}^2+2\text{as}$

${\text{v}}^2=2\times \text{g}\times 16$  

Since $25\%$ kinetic energy is lost in collision, the energy left in the body is$75\%$ therefore$\frac{75}{100}\times \frac{1}{2}{\text{mv}}^{\text{2}}=\text{mgh}$  

Where ‘h’ is the distance to which it rebounds $\left(\text{h}=12\text{m}\right)$

$\frac{75}{100}\times \frac{1}{2}{\text{mu}}^{\text{2}}=\text{mgh}$

$\frac{75}{100}\times \frac{1}{2}\times 2\times \text{g}\times 16=\text{g}\times \text{h}$

$\frac{75}{100}\times 16=\text{h}$

$\text{12}=\text{h}$

$\text{h}=\text{12m}$