The dimensions are $[m]=\mathsf M$, $[\alpha]=\mathsf M\mathsf T^{-1}$, $[u_0]=\mathsf L\mathsf T^{-1}$, and $[g]=\mathsf L\mathsf T^{-2}$. Under the stated model, $m,\alpha,u_0,g$ are the only parameters determining the ascent <time>. If $m^p\alpha^q u_0^r g^s$ is dimensionless, then $p+q=0$, $r+s=0$, and $-q-r-2s=0$, leaving one independent <dimensionless variable>. One convenient choice is
$$
\boxed{\lambda=\frac{\alpha u_0}{mg},\qquad T=\frac m\alpha f(\lambda).}
$$
Thus <dimensional analysis> fixes the form, but <Newton's second law> must determine the <function> $f$.

Take upward <velocity> as positive. Throughout ascent, <Newton's second law> with <linear drag> gives $m\dot v=-mg-\alpha v$. An <integrating factor> or direct solution of this <linear ordinary differential equation> gives
$$
v(t)=\left(u_0+\frac{mg}{\alpha}\right)e^{-\alpha t/m}-\frac{mg}{\alpha}.
$$
The first zero of this <velocity> is the highest point, so
$$
\boxed{f(\lambda)=\log(1+\lambda),\qquad T=\frac m\alpha\log\left(1+\frac{\alpha u_0}{mg}\right).}
$$
This <vertical ascent under linear drag> has $T<u_0/g$ for $u_0>0$, since $\log(1+\lambda)<\lambda$. The limit $\alpha\to0$ recovers the no-drag ascent <time> $u_0/g$; $u_0=0$ gives $T=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 4A / Solution

{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 $u_0/g$.

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 Incoming links: Vertical ascent under linear drag