Question

A spherical raindrop 3.1 mm in diameter falls through a vertical distance of 3550 m. Take the cross-sectional area of a raindrop = πr², 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 3550 m in the absence of air drag.

(b) What would its speed be at the end of 3550 m when there is air drag?

Answer 1

According to the given problem,

Volume of spherical raindrop = 4/3π r³ = 0.0156 cm³ and mass = 0.0156 gm

a) Speed in the absence of air drag.

S = 1/2 g t²

t² = 2 S / g = 3550 * 2 / 9.81 = 723.75

t = 26.9 s

v = g * t = 9.81 * 26.9 = 263.9 m/s

v = 264 m/s

b)

V_t = √2 m g / Cρ_air A
A = πr² = π(1.55*10^-3)² = 7.55*10^-6 m²
m = 0.0156 gm = 0.0156*10^-3 kg
g = 9.81
C = 0.45
ρ_air = 1.2 kg /m³

v_t = 8.66 m/s.

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