Question

A spherical raindrop 1.9 mm in diameter falls through a vertical distance of 4150 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 4150 m in the absence of air drag.
_________ m/s

(b) What would its speed be at the end of 4150 m when there is air drag? (Note that the raindrop will reach terminal velocity after falling about 30 m.)
_________ m/s

Answer 1

volum spherical raindrop = 4/3 π r^3 = 4/3 * π * 0.095 ^3 = 3.5913*10^3 cm3
mass = 0.01277 gram

a ) speed in the absence of air drag.
e = 1/2 g t^2 ---- t^2 = 2 e / g = 4150 * 2 / 9.8 = 846.93 ---- t = 29.1022 s
v= g * t = 9.8 * 29.1022 = 285.2015 m/s

b)
V terminal = √ 2 m g / C * ρ air * A
A = π r^2 =( 0.95 10^-3) ^2 * π = 2.8352* 10^-6 m2
m = 0.01277 gram = 0.01277 10^-3 kg
g = 9.8
C = 0.45
ρ air = 1.2 kg /m3

Vt = √ 2 * 0.01277 10^-3 * 9.8 / 0.45 * 1.2 * 2.28532* 10^-6
Vt = 14.2414 m/s