Descent from rest under quadratic drag ($v(t)=\sqrt{mg/\alpha }\tanh (\sqrt{\alpha g/m}t)$)

For positive downward speed, gravity and quadratic drag give $m\dot{v}=mg-\alpha v^2$, $v(0)=0$. Separation and the integral of $(1-s^2{)}^{-1}$ give the displayed approach to terminal velocity. The fallen distance is $(m/\alpha )\log \cosh (\sqrt{\alpha g/m}t)$. The limit as $\alpha \to 0$ recovers $v=gt$ and distance $g{t}^2/2$.