[bbrizzi]

The air pressure is proportional to either v or v² depending on the speed and atmosphere.

For a friction force Fr, the terminal velocity is reached when the acceleration becomes null:

mg + Fr = 0

mg - k* Vterm² = 0

k, the friction coefficient depends on different parameters: http://en.wikipedia.org/wiki/Drag_equation

So in the end, the terminal velocity will depend on:

- The guy's weight (contrary to a free-fall's acceleration which is independent of mass)

- The "contact surface" between the body and the atmosphere.

- The atmosphere's density (which is lower than on Earth's surface).

The guy aims at reaching Mach 1. Note that due to the lower atmospheric pressure at this altitude, Mach 1 is a bit smaller than it is on the Earth's surface ( 301 m/s at 29 000 meters and -48 degrees C compared to 340 m/s at sea level ).

Finally, the term "free fall" is not appropriate as the definition of a free fall is "any motion of a body where its weight is the only force acting upon it."( http://en.wikipedia.org/wiki/Free_fall ) This does not take into account the drag which is not negligible here.

[lutusp]

> The air pressure is proportional to either v or v² depending on the speed and atmosphere.

1. I think you mean "air resistance". Yes?

2. If so, then no, air resistance transitions from (linear) Stokes drag at low velocities to (aptly named) quadratic drag at higher velocities, as a function of velocity, but not a function of air pressure. So not "either v or v^2", but a combination of the two factors.

http://en.wikipedia.org/wiki/Drag_(physics)

[davidw]

It made my day to have my question answered by an honest-to-goodness rocket scientist. Thank you!

I think the numbers that best sum up the situation are provided by yardie and blaze33 though. I knew that stuff coming from space was obviously going to be moving faster than the guy's 0 velocity, but had no idea about the order of magnitude.

[bbrizzi]

Yes I did mean that, thanks for correcting me. I was never very good at fluid dynamics :)