Question

A body is projected vertically up with certain velocity. At point $P$ in its path, the ratio of its potential energy to kinetic energy is $36:64$. The ratio of velocity of projection to velocity at $P$ is

• A. $\frac{5}{4}$
• B. $\frac{4}{3}$
• C. $\frac{9}{4}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $\frac{5}{4}$


Given,
ratio of $PE$ to $KE$ at height $h$ is
$\frac{P{E}_h}{K{E}_h}=\frac{36}{64}$
$\frac{mgh}{\frac{1}{2}m{v}^2}=\frac{36}{64}$
$\Rightarrow v^2=\frac{32}{9}gh$
Also,
$v^2=u^2-2gh$ $(\because gh=\frac{9}{32}{v}^2)$
$\Rightarrow v^2=u^2-\frac{9}{16}{v}^2$
$\Rightarrow \frac{u^2}{v^2}=\frac{25}{16}$
$\Rightarrow \frac{u}{v}=\frac{5}{4}$.