Question

A particle us released from a height $H$. At certain height its kinetic energy is two times its potential energy. Height and speed of particle at that instant are

• A. $\frac{H}{3}.\sqrt{\frac{2gH}{3}}$
• B. $\frac{H}{3}.2\sqrt{\frac{gH}{3}}$
• C. $\frac{2H}{3}.\sqrt{\frac{2gH}{3}}$
• D. $\frac{H}{3}.\sqrt{2gH}$

Answer 1

The correct option is A $\frac{H}{3}.\sqrt{\frac{2gH}{3}}$

Total mechanical energy $=mgH$

Given $\frac{PE}{KE}=\frac{1}{2}$

$\therefore KE+PE=mgH$

$2PE+PE=mgH$

$\therefore PE=\frac{1}{3}mgH=mg\left(\frac{H}{3}\right)$

$\therefore$ height from ground at this instant is $\frac{H}{3}$.

$\therefore KE=\frac{2mgH}{3}$

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

$\therefore v=2\sqrt{\frac{9H}{3}}$ is the velocity at that instant.