= Solution
Measure displacement downward from the release point and take downward speed $v\geq0$. <Newton's second law> with <quadratic drag> gives
$$
m\frac{dv}{dt}=mg-\gamma v^2,
\qquad v(0)=0.
$$
Writing
$$
V=\sqrt{\frac{mg}{\gamma}},
\qquad k=\sqrt{\frac{g\gamma}{m}},
$$
separation of variables gives the <terminal velocity> solution
$$
v(t)=V\tanh(kt).
$$
The distance fallen by time $t$ is therefore
$$
y(t)=\int_0^tv(s)ds
=\frac{m}{\gamma}\log\cosh(kt).
$$
Setting $y(t)=h$ and solving for $t$ yields
$$
\boxed{t=\sqrt{\frac{m}{g\gamma}}\,
\operatorname{arcosh}\!\left(e^{\gamma h/m}\right)}.
$$
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