Question

Calculate the velocity a spherical rain drop would achieve falling (taking downward as positive) from 4.2 km in the following situations.

h = 4.2 km
l = 4.8 mm
d = 1.23 kg/m³

^a Calculate the velocity in the absence of air drag in m/s.

b  Calculate the velocity with air drag in m/s. Take the size across of the drop to be 4.8 mm, the density of air to be 1.23 kg/m³, the density of water to be 1000 kg/m³, the surface area to be πr², and the drag coefficient to be 1.0.

Answer 1

Part A. In the absence of air drag,

Using 3rd kinematic equation in vertical direction:

V^2 = U^2 + 2*a*h

U = Initial velocity of drop = 0 m/s

V = final velocity of drop = ?

a = acceleration due to gravity = g = 9.81 m/s^2

h = vertical displacement = 4.2 km = 4200 m

So,

V = sqrt (0^2 + 2*9.81*4200)

V = 287.1 m/s = final velocity of drop

Part B. When the air drag is present, then drop will reach it's terminal velocity.

terminal velocity of spherical drop will be given by:

Vt = sqrt (2*m*g/(Cd*rho*A))

m = mass of drop = Volume*density = (4/3)*pi*(2.4*10^-3)^3*1000 = 5.79*10^-5 kg

Cd = drag coefficient of sphere = 1.0

rho = density of air = 1.23 kg/m^3

A = Area of drop = pi*(2.4*10^-3)^2

So,

Vt = sqrt [2*5.79*10^-5*9.81/(1.0*1.23*pi*(2.4*10^-3)^2)]

Vt = 7.14 m/s = final speed of drop

Let me know if you've any query.