Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2019 / 4 / c

Neglecting $r_0$, the no-settling trajectory is $h_r=Ut$, while the settling trajectory is a downward-curving parabola, $h_r=Ut-k_{r}D{t}^2$. For the given coefficients $k_{r}D=3.0\times 10^{-5}\mathrm{m}{\mathrm{s}}^{-2}$. At $2000\mathrm{s}$,The radius is then $r=\sqrt{2Dt}=31.6\mu \mathrm{m}$, with terminal speed $0.12\mathrm{m}{\mathrm{s}}^{-1}$, still much less than the updraft. Such droplets are carried almost to cloud top rather than falling as rain. Condensation alone would take about $2\times 10^4\mathrm{s}$ 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.