Question

A spherical raindrop 2.7 mm in diameter falls through a vertical distance of 5000 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 5000 m in the absence of air drag.
313.04 m/s (correct)

(b) What would its speed be at the end of 5000 m when there is air drag? (Note that the raindrop will reach terminal velocity after falling about 30 m.)
  
_______ m/s

Answer 1

Diameter of spherical raindrop = d = 2.7 mm = 2.7 x 10^-3 m

Radius of spherical raindrop = r = d/2 = 1.35 x 10^-3 m

Vertical distance of fall = H = 5000 m

Cross sectional area of raindrop = A = r²

A = (1.35x10^-3)²

A = 5.72 x 10^-6 m²

Drag coefficient = C_d = 0.45

Density of water = _w = 1000 kg/m³

Density of air = _a = 1.2 kg/m³

Gravitational acceleration = g = 9.8 m/s²

Part (a)

Air drag is absent.

In this case the raindrop will be in freefall with no air resistance.

Initial velocity of raindrop = V₁ = 0 m/s

Final velocity of raindrop = V₂

(V₂)² = (V₁)² + 2gH

(V₂)² = 0² + 2x9.8x5000

(V₂)² = 98000

V₂ = 313.05 m/s

Part (b)

Air drag is present.

The raindrop reaches terminal velocity after falling about 30m that is it will continue falling with terminal velocity after that.

Hence the velocity of hthe raindrop at the end of 5000m is the terminal velocity of the raindrop

Terminal velocity of raindrop = V_t

For terminal velocity

(4/3)x1000x1.35x10^-3x9.8 = (1/2)x0.45x1.2xV_t²

V_t² = 65.33

V_t = 8.08 m/s

Speed of the raindrop at the end of 5000m when there is air drag is 8.08 m/s