I don’t think it will make it that much more complicated, but then again, MikeS is doing the heavy lifting here.
Great, the pressure’s on…
First, wind resistance may very well be a factor. Unfortunately, I’m going to have to ignore it, since it’s impossible to know what the magnitude of its effects would be. My rough estimate is that it might change things by a factor of two or three, but that’s really a WAG. If anyone wants to give me a rough analytic form to work with for the wind resistance (force as a function of velocity), then I might be able to modify things.
Here’s the result I get when I work through through this. We’ll denote q[sub]1[/sub] the mass loss due to shell casings, q[sub]2[/sub] the mass loss rate due to combustion products, and q[sub]3[/sub] the rate due to the bullets. Let u[sub]1[/sub], u[sub]2[/sub], u[sub]3[/sub] be the downward ejection velocity of each item, respectively. Then we end up with a very similar result to last time:
v(t) = gt  (P/q) ln(1  q t/m[sub]0[/sub])
where I’ve redefined some of my notation: P = q[sub]1[/sub] u[sub]1[/sub] + q[sub]2[/sub] u[sub]2[/sub] + q[sub]3[/sub] u[sub]3[/sub], and q = q[sub]1[/sub] + q[sub]2[/sub] + q[sub]3[/sub].
Note that the form of this motion is similar to the previous derivation, and thus the basic conclusion above still holds: the guy will be unable to stop himself unless his mass is below a certain value. In this case, M < P / g.
So what are the numbers? Well, we already know q[sub]3[/sub] and u[sub]3[/sub] from below. q[sub]2[/sub] is about 25 grains per round per gun, or about 40 gms/sec for both guns assuming the fire rate we’ve established below. u[sub]2[/sub] is unknown, but let’s assume that it averages out to about half of u[sub]3[/sub]. q[sub]1[/sub] is about 170 gms/sec from both guns, and we’ll assume that u[sub]1[/sub] = 0, i.e. the casings are not ejected with an appreciable vertical velocity (compared to the bullet and the combustion products, at least.)
Putting it all together, we find that :::taps away furiously on his calculator::: M < 14.8 kg, or about 33 pounds. I’m afraid that our intrepid hero is out of luck.
:::goes to massage his temples:::
Hee!
A factor that varies at roughly the cube of velocity, and limits a falling human to ~60 meters per second.
Humans have such a complicated shape that I think you’d actually have to measure things to find out how wind affects their acceleration.


 Hmmm, gonna need a couple M249s here. Maybe miniguns…
~
 Hmmm, gonna need a couple M249s here. Maybe miniguns…

Since it isn’t an elastic collision KE isn’t conserved. But p always is. Note that this answer would mean that a single bullet would impart ~3mph to the man. Enough to knock you over with a single bullet.
If you conserve momentum, it comes out as more like 686 bullets (if the other numbers are right and I got my math right).