A body is released from a height of 30   m vertically downwards. The speed of the body at which…

Suppose at velocity $v$,potential energy of body is twice that of kinetic energy.

Using energy conservation-

$mgH=\left(K{E}_f+P{E}_f\right) \dots \left(1\right)$

Since $P{E}_f=2\times K{E}_f$

Put this value in equation $\left(1\right)-$

$mgH=3\times K{E}_f$

$\Rightarrow mgH=3\times \frac{1}{2}m{v}^2$

$\Rightarrow v=\sqrt{\frac{2}{3}gH}$

$\Rightarrow v=10\sqrt{2} \mathrm{m} {\mathrm{s}}^{-1}$