Q. A particle is released from a height h, At a certain height; its KE is two times its potential energy. 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) $\frac{h}{3},\sqrt{2gh}$
Explanation
(b) Total mechanical energy= mgh
As, $\frac{KE}{PE} = \frac{2}{1}$
$KE = \frac{2}{3}mgh$
and $PE = \frac{1}{3}mgh$
Height from the ground at this instant,
$h^{,} = \frac{h}{3}\ and\ speed\ of\ particle\ at\ this\ instant,$
$v = \sqrt{2g\left( h - h^{'} \right)}$
$= \sqrt{2g\left( \frac{2h}{3}\ \right)}$
$= 2\sqrt{\frac{gh}{3}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.