How to calculate air resistance of penny dropped from Empire State Building? If a penny is dropped from the Empire State Building, then its speed, without taking air resistance into consideration, is
$\sqrt{\left(32\frac{\textrm{ft}}{\textrm{s}^2}\right)(1454\textrm{ ft})}=216.3\frac{\textrm{ft}}{\textrm{s}}$, or $147.5\frac{\textrm{mi}}{\textrm{h}}$.
What would the calculation be to take into account air resistance?
 A: As Brandon says in his comment, the terminal velocity of a penny is difficult to calculate because its passage through the air is not only turbulent but lacking the symmetry that makes approximate calculations possible for e.g. spheres.
Under these circumstances we hedge physicists resort to experiment, and this is exactly what Myth Busters did in 2003. The terminal velocity turns out to be 65 mph. This sounds fast, but because the penny is light its energy is too low to do any harm. From my memory of the programme one of the presenters actually had a penny fired at him at 65 mph to simulate the impact (now that's dedication to science :-) and while it hurt, no damage was done.
A: To answer your question you need to find at what speed the force of gravity is counteracted by the air resistance.
$$ m g = F_{\rm air}$$
For steady flow over a blunt object the air resistance is
$$ F_{\rm air}  = \frac{1}{2} \rho A C_d v^2 $$
Given some assumptions on the density of air, and coefficient of drag you can estimate the speed $v$.
If the coin is spinning things are more complicated because the frontal area $A$ and coefficient of drag changes with time and an average needs to be calculated.
