1. Initial setup

A particle is released from rest from height $S$ above the earth's surface.

• Initial potential energy: $U_i = mgS$
• Initial kinetic energy: $K_i = 0$

So total mechanical energy is:

$E = mgS$

2. At a general height $h$ above the surface

Let at some instant the particle be at height $h$ above the surface.

Then,

• Potential energy: $U = mgh$
• Kinetic energy: since total energy is conserved, $K = mgS - mgh = mg(S-h)$

3. Given condition

It is given that kinetic energy is three times the potential energy:

$K = 3U$

Substitute the expressions:

$mg(S-h) = 3mgh$

Cancel $mg$:

$S-h = 3h$

$S = 4h$

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

So the height from the surface is:

$\boxed{\frac{S}{4}}$

4. Find the speed at that instant

Now kinetic energy at that point is:

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

Using

$K = \frac{1}{2}mv^2$

we get

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

Cancel $m$:

$\frac{v^2}{2} = \frac{3gS}{4}$

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

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

5. Match with options

Thus the required height and speed are:

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

This matches Option C.