(a) A sphere made of plastic has a density of 1.08 g/cm^3 and a radius of 7.50 cm. It falls through air of density 1.20 kg/m^3 and has a drag coefficient of 0.500. What is its terminal speed (in m/s)?

Question

(b) From what height (in m) would the sphere have to be dropped to reach this speed if it fell without air resistance?

Answer 1

Hi Jeremy A.,

At terminal velocity we have, F_net = mg - F_D = ma = 0 or F_D = mg, where mg is the weight of the object and F_D is the drag force.

The equation for drag force F_D = (1/2)*C*ρ*A*(v_T)², where C is the drag coefficient, ρ is the density of air, A is the cross sectional area of the object, and v_T is the terminal velocity.

Substituting we get mg = (1/2)*C*ρ*A*(v_T)², and some algebra v_T = [2mg/ρCA]^1/2.

m = Density*Volume = 1.08*(4/3)*π*(7.50)³ g [cm³*g/cm³] = 1908.52 g = 1.9085 kg

g = 9.81 m/s²

ρ = 1.20 kg/m³

C = .500

A = π*(7.5)² cm² = 176.715 cm² = .0177 m²

Dimensional analysis shows [(kg*m/s²)/(m²*kg/m³)]^1/2 = m/s.

My answer 59.4 m/s = v_T.

For minimum height in a vacuum use h = v²/(2g), which can be derived from:

v = v_o + at (with v_o = 0, and a = g) and y = y_o + v_ot + (1/2)at² (h = y - y_o , v_o = 0, and a = g).

h = (v_T)²/(2g) = 179.8 m

I hope this helps, Joe.