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It is a common observation that rain clouds can be at about a kilometer altitude above the ground.
$(a)$ If a rain drop falls from such a height freely under gravity, what will be its speed ? Also calculate in $km/h$ $(g = 10\, m/s^2).$
$(b)$ A typical rain drop is about $4 \,mm$ diameter. Momentum is mass $\times $ speed in magnitude. Estimate its momentum when it hits ground.
$(c)$ Estimate the time required to flatten the drop.
$(d)$ Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
$(e)$ Estimate the order of magnitude force on umbrella. Typical lateral separation between two rain drops is $5\, cm$.
(Assume that umbrella is circular and has a diameter of $1\, m$ and cloth is not pierced through.)
Solution
$\text { Given, } h =1 \mathrm{~km}=1000 \mathrm{~m}$
$g=10 \mathrm{~m} / \mathrm{s}^{2}$
$\text { (a) } \text { Speed of drop in free fall, }$
$v^{2}=2 \mathrm{gh}$
$\therefore v=\sqrt{2 g h}$
$(b)$ Diameter of the drop,
$d=4 \mathrm{~mm}$ $\therefore$ Radius of the drop, $r=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$
Mass of a rain drop,
$m =\rho \times v$
$=\frac{4}{3} \pi r^{3} \rho$
$=\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3}\left(\because \rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}\right)$
$\approx 3.4 \times 10^{-5} \mathrm{~kg}$
Momentum of the rain drop,
$=p=m v$
$=3.4 \times 10^{-5} \times 100 \sqrt{2}$
$=4.7 \times 10^{-3} \mathrm{kgm} / \mathrm{s}$
$\approx 5 \times 10^{-3} \mathrm{kgm} / \mathrm{s}$
he diameter near the ground,
$t=\frac{d}{v}=\frac{4 \times 10^{-3}}{100 \sqrt{2}}=0.028 \times 10^{-3} \mathrm{~s}$
$=2.8 \times 10^{-5} \mathrm{~s} \approx 30 \mathrm{~ms}$
$(d)$ Force exerted by a rain drop,
$\mathrm{F}=\frac{\text { Change in momentum }}{\text { Time }}=\frac{\Delta p}{t}=\frac{p-0}{t}$
$=\frac{4.7 \times 10^{-3}}{2.8 \times 10^{-5}} \approx 168 \mathrm{~N}$
