Let required height of body is y. When body from rest falls through height (S - y), then under constant acceleration $v^2 = 0^2 + 2g(S - y) \implies v = \sqrt{2g(S - y)}$. When body is at height y above ground, potential energy $U = mgy$. As per given condition kinetic energy $K = 3U \implies \frac{1}{2}m(v)^2 = 3 \times mg(y) \implies \frac{1}{2} \times m \times 2g(S - y) = 3 \times mgy \implies S - y = 3y \implies y = \frac{S}{4}$. Then $v = \sqrt{2g(S - \frac{S}{4})} = \sqrt{\frac{3gS}{2}}$.