Question

A spherical raindrop 1.9 mm in diameter falls through a vertical distance of 5000 m. Take the cross-sectional area of a raindrop = ?r², drag coefficient = 0.45, density of water to be 1000 kg/m³, and density of air to be 1.2 kg/m³.

Calculate the speed a spherical raindrop would achieve falling from 5000 m in the absence of air drag.

What would its speed be at the end of 5000 m when there is air drag?

Answer 1

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Q) A spherical raindrop 2.9 mm in diameter falls through a vertical distance of 3850 m.?

Take the cross-sectional area of a raindrop = ?r2, drag coefficient = 0.45, density of water to be 1000 kg/m3, and density of air to be 1.2 kg/m3

(a) Calculate the speed a spherical raindrop would achieve falling from 3850 m in the absence of air drag.


(b) What would its speed be at the end of 3850 m when there is air drag?

ANS:

volumE OF spherical raindrop = 4/3 ? r^3 = 4/3 * ? * 0.145 ^3 = 0.012777 cm3
mass = 0.01277 gram

a ) speed in the absence of air drag.
e = 1/2 g t^2 ---- t^2 = 2 e / g = 3850 * 2 / 9.8 = 785.7 ---- t = 28 s
v= g * t = 9.8 * 28 = 274.7 m/s

b) V terminal = ? 2 m g / C * ? air * A
A = ? r^2 =( 1.45 10^-3) ^2 * ? = 2.1 10^-6 * ? = 6.6052 10^-6 m2
m = 0.01277 gram = 0.01277 10^-3 kg
g = 9.8
C = 0.45
? air = 1.2 kg /m3

Vt = ? 2 * 0.01277 10^-3 * 9.8 / 0.45 * 1.2 * 6.6052 10^-6 = 70.1725
Vt = 8.377 m/s