We can realize the situation as shown.  Let at point $C$ distance $x$ from highest point $A$, the particle's kinetic energy is three times its potential energy.

Velocity $C,$

$v^2=0+2gx$

$Or v^2=2gx$.....(i)

Potential energy at ,$C=mg \left(S-x\right) \dots ..\left(ii\right)$




At Point C,

Kinetic energy $=3\times$ potential energy

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

or $x=3S-3x$

or $4x=3S$

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

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

Therefore, from Eq.(i)

${ v}^2=2g\times \frac{3}{4}S$

Or ${ v}^2=\frac{3}{2}gS or v=\sqrt{\frac{3}{2}gS}$

Height of the particle from the ground

$= S-x=S-\frac{3}{4}S=\frac{s}{4}$