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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

A particle is released from height H above the ground ....

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)