Question

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

1. 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/s²)
2. A typical rain drop is about 4mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.
3. Estimate the time required to flatten the drop.
4. Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
5. 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 !!)

Answer 1

Given, height (h) = 1 km = 100 m

g = 10 m/s²

a. Velocity attained by the raindrop in freely falling through a height h.

v = `sqrt(2gh)`

= `sqrt(2 xx 10 xx 1000)`

= `100sqrt(2)` m/s

= `100sqrt(2) xx (60 xx 60)/1000` km/h

= `360sqrt(2)` km/h ≈ 510 km/h

b. Diameter of the drop (d) = 2r = 4 mm

∴ Radius of the drop (r) = 2 mm = 2 × 10^–3 m

Mass of a raindrop (m) = V × p

= `4/3 pir^3p`

= `4/3 xx 22/7 xx (2 xx 10^-3)^3 xx 10^3`  .....(∵ Density of water = 10³ kg/m³)

= 3.4 × 10^–5 kg

The momentum of the raindrop (p) = mv

= `3.4 xx 10^-5 xx 100sqrt(2)`

= `4.7 xx 10^-3` kg-m/s

≈ `5 xx 10^-3` kg-m/s

c. Time required to fatten the drop = time taken by the drop to travel the distance equal to the diameter of the drop near the ground

`t = d/v`

= `(4 xx 10^3)/(100 sqrt(2))`

= ``0.028 xx 10^-3` s

= `2.8 xx 10^-5` s ≈ 30 ms

d. Force exerted by a raindrop

F = `"Change in momentum"/"Time"`

= `(p - 0)/t`

= `(4.7 xx 10^-3)/(2.8 xx 10^-5)` ≈ 168 N

e. Radius of the umbrella (R) = `1/2` m

∴ Area of the umbrella (A) = `piR^2`

= `22/7 xx (1/2)^2`

= `22/28`

= `11/4 ≈ 0.8` m²

Number of drops striking the umbrella simultaneously with an average separation of 5 cm = `5 xx 10^-2` m

= `0.8/(5 xx 10^-2)^2` = 320

∴ Net force exerted on umbrella = 320 × 168 = 53760 N ≈ 54000 N.