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

Velocity at point C, $v^2=0+2gx\text{ Or }{v}^2=2gx$

Potential energy at C, $PE=mg(S-x)$

At point C, $KE=3PE$

$\Rightarrow \frac{1}{2}m\times 2gx=3\times mg(S-x)$

$\Rightarrow x=\frac{3}{4}S$

Height from the ground is $=S-x=S-\frac{3}{4}S=\frac{S}{4}$

Spee of the particle at C, $v^2=2gx=2g\frac{3S}{4}$

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