Vertical ascent under linear drag
= Vertical ascent under linear drag
{title2=$T=(m/\alpha)\log(1+\alpha u_0/(mg))$}
For constant downward gravitational <acceleration> $g$ and <linear drag> $-\alpha v$, the upward <velocity> solves $m\dot v=-mg-\alpha v$ with $v(0)=u_0\ge0$. Its first zero occurs at the displayed <time>. The dimensionless control variable is $\alpha u_0/(mg)$, and the limit $\alpha\to0$ gives $u_0/g$.