A particle is released freely from a height $H$. At a certain height, its kinetic energy is two times of its…
A particle is released freely from a height $H$. At a certain height, its kinetic energy is two times of its potential energy. Then, the height and the speed of the particle at that instant are respectively
( $g=$ acceleration due to gravity)
$\frac{H}{3}, \sqrt{\frac{2 g H}{3}}$
$\frac{H}{3}, 2 \sqrt{\frac{g H}{3}}$
$\frac{2 H}{3}, \sqrt{\frac{2 g H}{3}}$
$\frac{H}{3}, \sqrt{2 g H}$
Solution
If particle falls by a distance $x$, then
$
\begin{aligned}
\mathrm{KE}=\frac{1}{2} m v^2 & =\frac{1}{2} m(2 g x)=m g x \\
\mathrm{PE} & =m g(H-x)
\end{aligned}
$
As, $\mathrm{KE}=2(\mathrm{PE}) \Rightarrow m g x=2 m g(H-x) \Rightarrow x=\frac{2 H}{3}$
So, height of particle is $H-x=\frac{H}{3}$.
Speed of particle at $\frac{H}{3}$ distance $=\sqrt{2 g\left(\frac{2 H}{3}\right)}=2 \sqrt{\frac{g H}{3}}$