Raindrops

When the teardrop has radius $r$ and is travelling downwards with speed $v$ , it has mass $m = \tfrac43\pi\rho r^3$ (where $\rho$ is the density of water). The rate at which it increases in mass is equal to its cross-sectional area times its speed times the density of water, so that $\dot{m} \; = \; \pi r^2 \rho v$ and hence $v \; = \; 4\dot{r}$ Newton's Second Law tells us that $\frac{d}{dt}(mv) \; = \; mg$ and so $\begin{aligned} \tfrac{d}{dt}(r^3v) & = \; r^3g \\ \tfrac{d}{dt}(r^3\dot{r}) & = \; \tfrac14r^3g \\ r^3\ddot{r} + 3r^2(\dot{r})^2 & = \; \tfrac14r^3g \\ \tfrac{d}{dt}\left[\tfrac12r^6(\dot{r})^2\right] & = \; \tfrac14r^6g \\ \tfrac12r^6(\dot{r})^2 & = \; \tfrac{1}{28}r^7g + c \end{aligned}$ Since $\dot{r}$ should remain finite as $r \to 0$ , we deduce that $c=0$ , and hence $\tfrac12(\dot{r})^2 \; = \; \tfrac{1}{28}rg$ so that $\ddot{r} = \tfrac{1}{28}g$ , and hence $\dot{v} = 4\ddot{r} = \boxed{\tfrac17g}$ .