= Solution
During $dt$, the remaining rocket has <mass> $M+dM$ and <velocity> $v+dv$, while the expelled positive <mass> $-dM$ has <velocity> $v-u$ to first order. Subtract the initial <momentum> $Mv$ from their final <momentum> to obtain $M\,dv+u\,dM$. Equating this to the external <impulse> gives the <rocket equation>
$$
F=M\dot v+u\dot M.
$$
With $F=-Mg-kMv$ and $\dot M/M=-b$, this reduces to $\dot v+kv=ub-g$. Therefore, for initial <velocity> $v_0$ and $k>0$,
$$
v(t)=\frac{ub-g}{k}+\left(v_0-\frac{ub-g}{k}\right)e^{-kt}.
$$
For a launch from rest,
$$
\boxed{v(t)=\frac{ub-g}{k}(1-e^{-kt}),\qquad b>g/u\text{ for upward launch}.}
$$
The inequality makes the initial upward <acceleration> positive. Equality gives no upward launch from rest. In the zero-drag limit, $v=(ub-g)t$. These expressions apply while the prescribed burn law and constant-$g$ approximation remain valid.
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