Q.It is a common observation that rain clouds can be at about a kilometre altitude above the ground.

Rain Drop Impact Analysis

(a) Speed of the drop

A drop falling freely from height $h = 1000\ \text{m}$ under gravity ($g=10\ \text{m/s}^2$, $u=0$): …

(a) Free-fall speed from 1 km

Using $v^2 = u^2 + 2gh$ with $u=0$, $h = 1000\ \text{m}$, $g = 10\ \text{m/s}^2$:

Converting: $141.4 \times 3.6 \approx 509\ \text{km/h}$.

This chapter works with free-fall kinematics only (no drag/terminal-velocity model), so this same speed $v \approx 141\ \text{m/s}$ is carried through every remaining part.

(b) Momentum of a 4 mm drop

Diameter $d = 4\ \text{mm}$, so radius $r = 2\ \text{mm} = 2\times10^{-3}\ \text{m}$. Treating the drop as a sphere of water, $\rho = 1000\ \text{kg/m}^3$:

(c) Time to flatten the drop

The drop is estimated to come to rest over a distance roughly equal to its own diameter $d = 4\times10^{-3}\ \text{m}$, travelling at (essentially) its impact speed $v \approx 141\ \text{m/s}$:

(d) Force from one drop

Force is the rate of change of momentum. The drop's momentum drops from $p$ to $0$ in the flattening time $\Delta t$:

This is far larger than the drop's weight ($mg \approx 3.35\times10^{-4}\ \text{N}$) because the stop happens so suddenly.

(e) Force on a 1 m umbrella

Umbrella area: $A = \pi r_{\text{um}}^2 = \pi (0.5)^2 \approx 0.785\ \text{m}^2$. …

Concept: Estimating Impact Force via the Work-Energy Theorem Instead of the Impulse-Momentum Theorem

Method: A Second, Independent Estimation Route (Work Done = Kinetic Energy Lost) — Cross-Checking the Order of Magnitude, Not Reproducing the Same Numbers

The stored answer estimates each drop's stopping force via the impulse-momentum theorem ($F=\Delta p/\Delta t$, using a separately estimated flattening time). This method instead estimates the same force via the work-energy theorem ($F\cdot d \approx \Delta KE$, using the stopping distance directly, with no separate time-estimate step) — a genuinely different physical route to an order-of-magnitude estimate, which (honestly) lands on a numerically different but consistently order-matching answer, exactly as expected for an "estimate the order of magnitude" question.

(a) Free-fall speed — necessarily the same starting point

(b) Momentum — likewise unavoidable groundwork

Mass of a $4\ \text{mm}$-diameter water drop ($r=2\times10^{-3}\ \text{m}$, $\rho=1000\ \text{kg/m}^3$): $m = \tfrac43\pi r^3\rho \approx 3.35\times10^{-5}\ \text{kg}$, so $p = mv \approx 4.7\times10^{-3}\ \text{kg·m/s}$.

(c)+(d) — the work-energy route, computed together (no separate time estimate needed first)

1. State the work-energy theorem for the stopping process: the net force doing work over the stopping distance removes all the drop's kinetic energy:

assuming (for an order-of-magnitude estimate) the stopping force is roughly constant over the flattening distance $d\approx4\ \text{mm}$ (the drop's own diameter).

2. Solve directly for the force — this is the estimate for (d), reached WITHOUT first estimating a time:

3. Only now, having $F$, back out the implied flattening time (c) from the impulse relation, as a consistency step rather than an input:

(e) Force on the umbrella — same simultaneous-impact count, new per-drop force

Umbrella area $A=\pi(0.5)^2\approx0.785\ \text{m}^2$; with drops spaced $\ell=5\ \text{cm}$ apart, $N = A/\ell^2 \approx 314$ drops striking simultaneously (same geometric estimate as before — this part of the reasoning doesn't depend on which force model was used for a single drop):

Why the two methods disagree by roughly a factor of 2 — and why that's expected, not a mistake …