The velocity of freely falling body changes as $g^{p} h^{q}$ where $g$ is acceleration due to gravit

Question

The velocity of freely falling body changes as $g^{p} h^{q}$ where $g$ is acceleration due to gravity and $h$ is the height. The value of $p$ and $q$ are\

Professor PC Thomas Classes · Answer

$\mathrm{v} \propto \mathrm{g}^{\mathrm{p}} \mathrm{h}^{\mathrm{q}}$\ velocity of a body when it falls through a distance $h$ is\ $v=\sqrt{2 g h}$\ (2)\ $=\sqrt{2} \mathrm{~g}^{1 / 2} \mathrm{~h}^{1 / 2}$\ Comparing $\mathrm{p}=\frac{1}{2} \quad \mathrm{q}=\frac{1}{2}$\