Question

A particle is 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. $\frac{H}{3}, \, \sqrt{\frac{2 g H}{3}}$
• B. $\frac{H}{3}$ , $2\sqrt{\frac{g H}{3}}$
• C. $\frac{2 H}{3}$ , $\sqrt{\frac{2 g H}{3}}$
• D. $\frac{H}{3}$ , $\sqrt{2 g H}$

Answer 1

Answer: (B)

$\therefore K+U=mgH$ & $K=2U$
$\therefore 2U+U=mgH$
$\Rightarrow U=\frac{m g H}{3} \, \Rightarrow mgh=\frac{m g H}{3} \, \Rightarrow \, h=H / 3$
$\therefore \frac{1}{2} \, m v^{2}=2U=\frac{2 m g H}{3} \, \Rightarrow v=2\sqrt{\frac{g H}{3}}$