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 calcualte in `km//h (g = 10 m//s^(2))`.
(b) A typical rain drop is about 4 mm diameter. Momentum is mass `xx` speed in magnitude. Estiamate its momentum when its hits ground.
(c) Estimate the time required to flatten the drop.
(d) Rate of change of momentum is force. Estiamate 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 drop is 5 cm.
(Assume that umbrealla is circular and has a diameter of 1 m and cloth is not peicreced through.)

Explanation

Given, height `(h) = 1 km = 1000 m`
`g = 10 m//s`
(a) velocity attained by the rain drop in freely falling through a height h.
`v = sqrt(2gh) = sqrt(2 xx 10 xx 1000) = 100sqrt(2) m//s`
`= 100 sqrt(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 xx 10^(-3) m`
Mass of rain drop `(m) = V xx rho`
`= 4/3 pir^(3)rho`
`= 4/3 xx 22/7 xx (2 xx 10^(-3))^(2) xx 10^(3)` , (`:.` Density of water `= 10^(3) kg//m^(3)`)
`= 3.4 xx 10^(-5) kg`
Momentum of the rain drop `(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 flatten 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))/(100sqrt(2)) = 0.028 xx 10^(-3) s`
`= 2.8 xx 10^(-5) s = 30s`
(d) Force exerted by a rain drop
`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) = 11/14 ~~ 0.8 m^(2)`
Number of drops striking the umbrella simultaneously with average separation of 5 cm `= 5 xx 10^(-2) m`
`= (0.8)/((5 xx 10^(-2))^(2)) = 320`
`:.` Net force exerted on umbrella `= 320 xx 168 = 53760 N = 54000 N`