A falling spherical object of mass $m$ with surface area $S$ and constant density $p$ experiences an atmospheric drag proportional to $Sv^2$, where $v$ is the velocity of the object.
How do you show that, when the object reaches its terminal velocity, the kinetic energy per unit of its surface area is proportional to $m^{2/3}$.
The kinetic energy is defined as the mass of the object multiplied by its velocity squared and the terminal velocity is reached when the falling object stops accelerating due to the air drag.
The following relationships can be established for the spherical object,
$$mg= Sv^2$$ $$S=4\pi r^2$$ $$m=\frac{4}{3}\pi r^3 p$$
where $r$ is the radius of the sphere and the first equation equalizes the gravitational force with the drag.
So, the kinetic energy per unit of surface area is
$$\frac{mv^2}{S}=\frac{m^2g}{S^2}=\frac{m^2g}{16\pi^2 r^4}=\frac{m^2g}{16\pi^2 \left(\frac{3m}{4\pi p} \right)^{4/3 }}$$ $$=\frac{m^{\frac{2}{3}}g}{\left(\frac{36\pi}{p^2} \right)^{\frac{2}{3}}}$$
which is proportional to $m^{2/3}$.