Descent from rest under quadratic drag
= Descent from rest under quadratic drag
{title2=$v(t)=\sqrt{mg/\alpha}\,\tanh(\sqrt{\alpha g/m}\,t)$}
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 $\alpha\to0$ recovers $v=gt$ and distance $gt^2/2$.