Vertical ascent under linear drag ($T=(m/\alpha )\log (1+\alpha u_0/(mg))$)

For constant downward gravitational acceleration $g$ and linear drag $-\alpha v$, the upward velocity solves $m\dot{v}=-mg-\alpha v$ with $v(0)=u_0\ge 0$. Its first zero occurs at the displayed time. The dimensionless control variable is $\alpha u_0/(mg)$, and the limit $\alpha \to 0$ gives $u_0/g$.