Using the equation of motion for velocity at height $ x $:

$ V^2 = 0 + 2g(S - x) $

Simplifies to:

$ V^2 = 2g(S - x) $

Potential Energy at height $ x $:

$ \text{Potential Energy} = mgx $

Relation between kinetic and potential energy:

Given that the kinetic energy is three times the potential energy,

$ mgx = 3 \times \frac{1}{2} mV^2 $

Simplifying gives:

$ gx = \frac{3}{2} \times 2g(S - x) $

$ gx = 3g(S - x) $

$ 4x = S $

Solving for $ x $:

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

Finding the speed $ V $ at this instant:

$ V = \sqrt{2g \left( S - \frac{S}{4} \right)} = \sqrt{2g \times \frac{3S}{4}} = \sqrt{\frac{3gS}{2}} $

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