= Solution
Introduce the <dimensionless quantities> $\tau=\mu t/r_0$ and $w=\mu v/(gr_0)$. Then
$$
w(\tau)=\frac14\left[(1+\tau)-(1+\tau)^{-3}\right].
$$
The <Taylor series> at $t=0$ gives
$$
\boxed{v(t)=gt-\frac{3g\mu}{2r_0}t^2+\frac{5g\mu^2}{2r_0^2}t^3+O(t^4),\qquad t\ll r_0/\mu.}
$$
Thus the initial slope is $g$, as for <free fall>. At late <times>,
$$
\boxed{v(t)=\frac{gt}{4}+\frac{gr_0}{4\mu}-\frac{gr_0^4}{4\mu(r_0+\mu t)^3}\sim\frac{gt}{4},\qquad t\gg r_0/\mu.}
$$
The oblique <asymptote> is $v=gt/4+gr_0/(4\mu)$, approached from below. Direct differentiation gives
$$
\dot v=\frac g4\left[1+3(1+\tau)^{-4}\right]>0,\qquad\ddot v=-\frac{3g\mu}{r_0}(1+\tau)^{-5}<0.
$$
The <speed> increases monotonically with decreasing slope, from $g$ towards $g/4$. It does not reach a <terminal velocity> in this drag-free model. Without <mass> accumulation, and still without <drag>, a drop released from rest has $v=gt$: its <acceleration> remains $g$, whereas the growing drop continually shares its <momentum> with newly acquired material.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022-q7-drop-speed.png]
{title=Growing-droplet <speed>, its early and late limits, and an explicitly assumed <quadratic drag> example}
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