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 [~~cm~~^{3}*g/~~cm~~^{3}] = 1908.52 g = 1.9085 kg

g = 9.81 m/s^{2}

ρ = 1.20 kg/m^{3}

C = .500

A = π*(7.5)^{2} cm^{2} = 176.715 cm^{2} = .0177 m^{2}

Dimensional analysis shows [(~~kg~~*m/s^{2})/(~~m~~^{2}*~~kg~~/m^{3})]^{1/2} = m/s.

My answer 59.4 m/s = v_{T}.

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

v = v_{o} + at (with v_{o} = 0, and a = g) and y = y_{o} + v_{o}t + (1/2)at^{2} (h = y - y_{o} , v_{o} = 0, and a = g).

h = (v_{T})^{2}/(2g) = 179.8 m

I hope this helps, Joe.