= Solution
Neglecting $r_0$, the no-settling trajectory is $h_r=Ut$, while the settling trajectory is a downward-curving parabola, $h_r=Ut-k_rDt^2$. For the given coefficients $k_rD=3.0\times10^{-5}\,\mathrm{m\,s^{-2}}$. At $2000\,\mathrm s$,
$$
\boxed{h_{\rm no\ fall}=2000\,\mathrm m,\qquad h_r=1880\,\mathrm m,\qquad\Delta h_r=120\,\mathrm m.}
$$
The radius is then $r=\sqrt{2Dt}=31.6\,\mu\mathrm m$, with terminal speed $0.12\,\mathrm{m\,s^{-1}}$, still much less than the updraft. Such droplets are carried almost to cloud top rather than falling as rain. Condensation alone would take about $2\times10^4\,\mathrm s$ to reach $100\,\mu\mathrm m$, much longer than the approximately $2000\,\mathrm s$ cloud transit time. It also reduces relative radius differences because $r^2$ grows at the same rate for every droplet, leaving inefficient differential gravitational collection. This is the warm-rain growth bottleneck, not a claim that condensation is absent.
\Image[/past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2019-cloud-trajectories.png]
{title=Condensational droplets rise through the cloud while rapidly collecting drops can turn and precipitate}
{description=Trajectories calculated from the two growth models in question 4; the collector curve is solid only while the drop remains within the cloud. The dashed extension below cloud base is a formal extrapolation.}
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