A particle is released from height H above the ground. At a certain instant its kinetic energy is three times its potential energy. The height h above the ground and the speed of the particle v at that instant are respectively:

(1) $\frac {H}{4},\frac {3gH}{2}$

(2) $\frac {H}{4},\frac {\sqrt {3gH}}{2}$

(3) $\frac {H}{2},\frac {\sqrt {3gH}}{2}$

(4) $\frac {H}{4},\sqrt {\frac {3gH}{2}}$

*Solution*

Using conservation of mechanical energy we have,

$mgH=KE+PE=4PE=4mgh$ (given, KE = 3PE)

$\Rightarrow h=\frac {H}{4}$

Now, $\frac {1}{2}mv^2=3mgh=3mg \frac {H}{4}$

$\Rightarrow v=\sqrt \frac {3gH}{2}$

Answer: (4)