Hi Jeremy A.,
At terminal velocity we have, Fnet = mg - FD = ma = 0 or FD = mg, where mg is the weight of the object and FD is the drag force.
The equation for drag force FD = (1/2)*C*ρ*A*(vT)2, where C is the drag coefficient, ρ is the density of air, A is the cross sectional area of the object, and vT is the terminal velocity.
Substituting we get mg = (1/2)*C*ρ*A*(vT)2, and some algebra vT = [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)/( m 2* kg/m 3)]1/2 = m/s.
My answer 59.4 m/s = vT.
For minimum height in a vacuum use h = v2/(2g), which can be derived from:
v = vo + at (with vo = 0, and a = g) and y = yo + vot + (1/2)at2 (h = y - yo , vo = 0, and a = g).
h = (vT)2/(2g) = 179.8 m
I hope this helps, Joe.