 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 $v(0)=v_0$: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.