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 :

To find the height from the surface of the Earth and the speed of the particle at the given instant, we can use the law of conservation of mechanical energy.

Let:
- $m$ be the mass of the particle.
- $g$ be the acceleration due to gravity.
- The surface of the Earth be the reference level where potential energy is zero ($U=0$).

Step 1: Calculate the total mechanical energy of the particle.
The particle is released from rest at a height $S$. Therefore, its initial velocity is zero, and its initial kinetic energy is also zero ($K_i=0$).
The initial potential energy of the particle at height $S$ is:

$U_i=mgS$

The total mechanical energy ($E$) of the system is the sum of its kinetic and potential energies:

$E=K_i+U_i=0+mgS=mgS$

Step 2: Find the height from the surface of the Earth.
Let $h$ be the height of the particle from the surface of the Earth at the instant when its kinetic energy ($K$) is three times its potential energy ($U$).
At height $h$:
- Potential energy is $U=mgh$
- Kinetic energy is given as $K=3U=3mgh$

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

$K+U=E$

Substitute the values of $K$, $U$, and $E$ into the equation:

$3mgh+mgh=mgS$

$4mgh=mgS$

Dividing both sides by $mg$:

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

Step 3: Find the speed of the particle.
Let $v$ be the speed of the particle at height $h$. The kinetic energy is expressed as:

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

Since $K=3mgh$ and we found $h=\frac{S}{4}$:

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

Equating the two expressions for kinetic energy:

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

Dividing both sides by $m$ and multiplying by 2:

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

Taking the square root of both sides:

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

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