> Clearly, he's falling from a different altitude than a rocket, and starting from 0, but what are the actual numbers like?

Without an atmosphere, the result would be deterministic and trivial to compute:

acceleration = f(t) = g

velocity = ∫ f(t) dt = g * t

position = ∫∫ f(t) dt = 1/2 g t^2

(The value of g changes slightly for this problem, in this altitude domain.)

But because of the atmosphere, the calculation of velocity as a function of altitude is much more difficult. And it turns out that the atmospheric pressure as a function of altitude is not trivially characterized. And at high altitudes, it's not even constant -- it depends on temperature, the position of the sun in the sky, even the time in the 11-year sunspot cycle.

The air resistance of a falling object is some constant k (based on the object's size and surface roughness) times the square of the velocity times the air pressure. But the air pressure is changing as the descent unfolds, so such a computation ultimately relies on a numerical solution to a differential equation.

This is why one doesn't see a trivial equation describing descent velocity for a skydiver. I have worked out skydiver velocity profiles for constant air pressure nearer the surface:

http://arachnoid.com/sage/terminal_velocity.html

But this equation, only an approximation at lower altitudes, is of no use at all for a problem like the Baumgartner jump.