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 $4\, mm$ in diameter. Momentum is mass $\times$ speed in magnitude. Estimate its momentum when it hits the 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 an umbrella. Typical lateral separation between two rain drops is $5\, cm$.
(Assume that the umbrella is circular and has a diameter of $1\, m$ and the cloth is not pierced through.)

(D) Given: $h = 1\, km = 1000\, m$,$g = 10\, m/s^2$,$\rho = 10^3\, kg/m^3$.
$(a)$ Using $v^2 = u^2 + 2gh$ with $u = 0$: $v = \sqrt{2 \times 10 \times 1000} = \sqrt{20000} \approx 141.4\, m/s$. In $km/h$: $141.4 \times (18/5) \approx 509\, km/h$.
$(b)$ Radius $r = 2\, mm = 2 \times 10^{-3}\, m$. Mass $m = \rho \times (4/3)\pi r^3 = 10^3 \times (4/3) \times 3.14 \times (2 \times 10^{-3})^3 \approx 3.35 \times 10^{-5}\, kg$. Momentum $p = mv = 3.35 \times 10^{-5} \times 141.4 \approx 4.74 \times 10^{-3}\, kg\cdot m/s$.
$(c)$ Time to flatten $t = d/v = (4 \times 10^{-3}) / 141.4 \approx 2.83 \times 10^{-5}\, s$.
$(d)$ Force $F = \Delta p / \Delta t = (4.74 \times 10^{-3}) / (2.83 \times 10^{-5}) \approx 167.5\, N$.
$(e)$ Area of umbrella $A = \pi R^2 = \pi (0.5)^2 \approx 0.785\, m^2$. Separation $s = 5\, cm = 0.05\, m$. Number of drops $N = A / s^2 = 0.785 / (0.05)^2 = 314$. Total force $F_{total} = N \times F \approx 314 \times 167.5 \approx 5.26 \times 10^4\, N$.