Explanation

(d) 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.
$Velocity\ at\ C,$
$v^{2} = 0 + 2gx$
$Or\ \ \ \ \ v^{2} = 2gx\ \ \ \ \ \ \ \ $…..(i)
Potential energy at C,$= mg\ (S - x)\ \ \ldots..(ii)$
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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
$= \ S - x = S - \frac{3}{4}S = \frac{s}{4}$