= Solution
Set $v=\dot z$. The second-order position equation becomes the first-order <linear ordinary differential equation> $\dot v+\gamma v=-g$. For $\gamma>0$, multiply by the <integrating factor> $e^{\gamma t}$ and impose $v(0)=v_0$:
$$
\boxed{v(t)=\left(v_0+\frac g\gamma\right)e^{-\gamma t}-\frac g\gamma.}
$$
The limiting <terminal velocity> is $-g/\gamma$, consistent with upward-positive position and downward <gravity>. The continuous zero-drag limit is $v_0-gt$; one must take that limit of the whole expression rather than set $\gamma=0$ in its two singular-looking terms separately.
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