Question

A particle is released from a height $H$. At a certain height, its kinetic energy is two times its potential energy.The height and speed of the 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 B $\frac{H}{3},2\sqrt{\frac{gH}{3}}$

Using law of conservation of energy,
$K_i+U_i=K_f+U_f$
let initial position be when particle is dropped from height $H$
$U_i=mgH$
$K_i=0$
Given $K_f=2U_f$
$K_i+U_i=K_f+U_f$
$mgH=3U_f$

$U_f=\frac{1}{3}mgH$ and $K_f=\frac{2}{3}mgH$
Now using $K_f=\frac{m{v}^2}{2}$
and speed of particle will be
$v=\sqrt{2g\left(\frac{2H}{3}\right)}=2\sqrt{\frac{gH}{3}}$
Now using $U_f=mgh$
$h=\frac{H}{3}$