A particle is released from height S from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively :

The correct answer is:
$\frac{S}{4},\sqrt{\frac{3gS}{2}}$

Step-by-step derivation:

1. Identify the Initial Mechanical Energy:
Let the surface of the Earth be the reference level where the potential energy is zero.
A particle of mass $m$ is released from rest from a height $S$ above the surface of the Earth.
Since it is released from rest, its initial velocity is zero, and thus its initial kinetic energy is zero.
The total mechanical energy ($E$) of the particle initially is purely potential energy:
$E=U_{\text{initial}}+K_{\text{initial}}=mgS+0=mgS$

2. Express Energy at height $h$:
Let $h$ be the height from the surface of the Earth at the instant of interest.
At this height $h$, the potential energy ($U$) is:
$U=mgh$
According to the problem description, at this instant, the kinetic energy ($K$) is three times its potential energy:
$K=3U=3mgh$

3. Calculate the Height $h$ using Conservation of Energy:
According to the law of conservation of mechanical energy, the total energy remains constant:
$E=K+U$
Substitute the expressions for $E$, $K$, and $U$:
$mgS=3mgh+mgh$
$mgS=4mgh$
Dividing both sides by $mg$ gives:
$S=4h$
$h=\frac{S}{4}$

4. Calculate the Speed $v$ at this instant:
The kinetic energy of the particle with speed $v$ is given by:
$K=\frac{1}{2}m{v}^2$
We already established that $K=3mgh$. Substitute $h=\frac{S}{4}$:
$K=3mg\left(\frac{S}{4}\right)=\frac{3mgS}{4}$
Equating the two expressions for kinetic energy:
$\frac{1}{2}m{v}^2=\frac{3mgS}{4}$
Cancel mass $m$ from both sides and multiply by 2:
$v^2=\frac{3gS}{2}$
Taking the square root:
$v=\sqrt{\frac{3gS}{2}}$

Thus, the height from the surface of Earth and the speed of the particle are respectively:
$\frac{S}{4}\text{and}\sqrt{\frac{3gS}{2}}$