A water drop of mass m = 0.1 g falls from rest from a height…

A water drop of mass $m = 0.1\ \mathrm{g}$ falls from rest from a height $H = 500\ \mathrm{m}$. It evaporates so that its mass decreases by $\Delta m = kt$, $k = 0.01\ \mathrm{g/s}$. At what height is its kinetic energy greatest? What fraction of its mass has it lost then, and what is that maximum kinetic energy? (Neglect air resistance; the evaporation does not push the drop.)

Explanation of Method

Maximum kinetic energy of an evaporating drop.

Step 1: Kinetic energy

The vapour leaves evenly in all directions, so it exerts no net force: the drop falls freely, $v = gt$.

\[ E_k(t) = \tfrac12\,(m - kt)\,g^2t^2. \]

Step 2: Maximum

\[ \frac{dE_k}{dt} = \tfrac12 g^2\,(2mt - 3kt^2) = 0, \]

\[ t_m = \frac{2m}{3k} = \frac{20}{3} \approx 6.67\ \mathrm{s}. \]

Step 3: Results

\[ \tfrac12 g t_m^2 = 218\ \mathrm{m}, \]

\[ \frac{\Delta m}{m} = \frac{kt_m}{m} = \frac23, \]

\[ v_m = gt_m = 65.4\ \mathrm{m/s}. \]

\[\begin{gathered} h \approx 282\ \mathrm{m}, \quad \frac{\Delta m}{m} \approx 67\% \\ E_{k,\max} \approx 0.071\ \mathrm{J} \end{gathered}\]

$E_k\ (\mathrm{J})$ $t\ (\mathrm{s})$ $0$ $2$ $4$ $8$ $10$ $0.071$ $6.67$ $0.071\ \mathrm{J}\text{ at 282 m}$

Kinetic energy of the evaporating drop. At first the growing speed wins; later the shrinking mass wins, and the energy falls to zero when the drop has evaporated after 10 s.

Result

\[\boxed{\,\begin{gathered} h \approx 282\ \mathrm{m}, \quad \frac{\Delta m}{m} \approx 67\% \\ E_{k,\max} \approx 0.071\ \mathrm{J} \end{gathered}\,}\]

Remark

Check. The original results are correct (height ≈ 280 m, more precisely 282 m).