If’n I also change the cross-sectional area and the weight, the kid+Hummer would fall about 500 metres
… but the Hummer would STILL be toast :^)

Now, if’n we’re to listen to good ol’ Newton, we’d write:
Force = (mass) (acceleration) or F = ma
and that gives
(Gravitational Force)-(Drag Force) = mdv/dt
and the Maximum Velocity (actually it’s the “speed”) is when
(Gravitational Force) = (Drag Force)
’cause that’s when dv/dt = 0 and the kid ain’t changing his speed.

So we got
mg (that’s Gravitationl Force) = k v^2 (that’s Drag Force)
assuming (Drag Force) is proportonal to velocity^2.
(Did I mention that it’s really the “speed”, not the “velocity”?)

If’n we didn’t take this short cut, we’d have to solve a differential equation, eh?
m dv/dt = m g – k v^2
which’d give us a tanh function
… and we don’t wanna do that!

So then we got: m g – k v^2 = 0
so: Maximum Velocity = SQRT[mg/k ]

Now Drag Force is like so:
Drag Force = k v^2 = (1/2) C p A v^2
where C depends upon the geometry of the falling kid
and p is the density of air (kg/m^3)
and A is the cross-sectional area of the falling kid (and NOT the surface area )
… and it’s in square metres.

So we now got:
k = (1/2) C p A
so
Max. Velocity=SQRT[mg/k ]=SQRT[2mg/C p A]
which turns out to be 36.15 m/s
- using Matthew’s numbers
- jest like on Matthew’s Neato Chart.

If he fell at that Maximum speed for 28.4 seconds, he’d have fallen over 1000 metres.

Note (from Matthew’s Neato Chart) how quickly the Maximum Velocity is reached.

Note, too, that if the kid fell over while driving a Hummer then the drag coefficient, C, would be 26.5
(instead of C = 1) and he’d fall only about 200 metres … but the Hummer would be toast.