Question

A particle is released from a height H and at certain height its kinetic energy is two times its potential energy, then the height and speed of the particle at that instant are

• A. $\frac{H}{3},\sqrt{\frac{4gH}{3}}$
• B. $\frac{2H}{3},\sqrt{\frac{2gH}{3}}$
• C. $\frac{2H}{3},\sqrt{\frac{2gH}{3}}$
• D. $\frac{H}{3},\sqrt{2gH}$

Answer 1

The correct option is A $\frac{H}{3},\sqrt{\frac{4gH}{3}}$

Total Energy T.E = mgH
Given K.E=2.PE at some height.
Conserving T.E
K.E + P.E = mgH
$\begin{array}{cc}3P.E=mgH& K.E=\frac{2}{3}mgH=\frac{1}{2}m{v}^2\\ P.E=mg\frac{H}{3}& v^2=\frac{4}{3}gH\\ height=\frac{H}{3}& v=\sqrt{\frac{4}{3}gH}\end{array}$