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 option is:

Step-by-Step Explanation:

1. Understanding the Initial State:
Let a particle of mass $m$ be released from rest at a height $S$ above the surface of the Earth.
At the maximum height $S$:
- Initial velocity $u=0$
- Potential Energy ($P{E}_{\text{initial}}$) = $mgS$
- Kinetic Energy ($K{E}_{\text{initial}}$) = $0$
- Total Mechanical Energy ($E$) = $mgS$

2. Finding the Height where $KE=3PE$:
Let $h$ be the height of the particle from the surface of the Earth at a given instant.
At height $h$:
- Potential Energy ($PE$) = $mgh$
- Kinetic Energy ($KE$) = $3\times PE=3mgh$

According to the law of conservation of mechanical energy, the total energy remains constant at any point:

$KE+PE=E$

Substituting the values in terms of $h$:

$3mgh+mgh=mgS$

$4mgh=mgS$

Dividing both sides by $mg$:

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

Thus, the height of the particle from the surface of the Earth at this instant is $\frac{S}{4}$.

3. Calculating the Speed at this Instant:
At height $h=\frac{S}{4}$, the kinetic energy is given by:

$KE=3mgh=3mg\left(\frac{S}{4}\right)=\frac{3mgS}{4}$

We also know that kinetic energy in terms of speed $v$ is:

$KE=\frac{1}{2}m{v}^2$

Equating the two expressions for kinetic energy:

$\frac{1}{2}m{v}^2=\frac{3mgS}{4}$

Canceling mass $m$ from both sides and multiplying by 2:

$v^2=\frac{3gS}{2}$

Taking the square root on both sides:

$v=\sqrt{\frac{3gS}{2}}$

Conclusion:
The height from the surface of Earth and the speed of the particle at that instant are respectively:

$\frac{S}{4}\text{, }\sqrt{\frac{3gS}{2}}$