The kinetic energy of freely falling body

$\text{K}\text{.E}=\frac{1}{2}{\text{mv}}^{\text{2}}\left[\because \text{v}=\text{at}\right]$

$\text{K}\text{.E}=\frac{1}{2}\text{m}{\text{g}}^2{\text{t}}^{\text{2}}$

$\text{K}\text{.E}\propto {\text{t}}^{\text{2}}$

$\text{K}\text{.E}=\frac{1}{2}\text{m}2{\text{gh v}}^2=\text{2gh}$

$\text{K}\text{.E}=\text{mgh }$

$\text{K}\text{.E}\propto \text{h}$