 Solution

Take upward velocity as positive. The linear drag force is $-mkv$, so it opposes the motion both during ascent and descent. Newton's second law gives $\dot{v}+kv=-g$. With $v(0)=u$, the integrating factor $e^{kt}$ yieldsThe terminal velocity is $-g/k$. The limit $k\to 0$ at fixed $t$ recovers $u-gt$; expanding the exponential is necessary to see the cancellation of the apparently singular terms.