Question

A spherical raindrop 2.9 mm in diameter falls through a vertical distance of 3950 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 (in m/s) a spherical raindrop would achieve falling from 3950 m in the absence of air drag.

278.222 m/s

(b)

What would its speed (in m/s) be at the end of 3950 m when there is air drag?

_______ m/s

Answer 1

Given that,

diameter = 2.9 mm

radius, r = D / 2 = 1.45 mm

h = 3950 m

drag coefficient, C = 0.45

(a)

ln absent of air drag,

From kinematic equation,

h = ut + (1/2)at^2

3950 = 0 + (1/2)9.8*t^2

t = 28.39 s

speed of spherical raindrop, v = u + at

v = 0 + 9.8*28.39

v = 278.24 m/s

(b)

Volume of raindrop,

V = (4/3)pi*r^3 = (4/3)*pi*(1.45*10^-3)³ = 0.0127*10^-6 m^3

Mass of drop = V* =  0.0127*10^-6*1000

m =  0.0127*10^-3 kg

ln presence of air drag,

Terminal velocity of drop,

v = sqrt (2mg / C*a*A)

v = sqrt [2*0.0127*10^-3*9.8 / 0.45*1.2*pi*(1.45*10^-3)²]

v = 8.37 m/s