If you want to take gravity into account, things get a little more complicated. Warning: semi-complicated math ahead.

This is essentially the rocket equation in a constant gravitational field. (Unfortunately, the equations on that page are broken, but I couldn’t find anything better on the web; I’m following my old intermediate mechanics book.) The end result is that the guy’s velocity as a function of time is given by

v(t) = -gt - u ln(1 - q t/m[sub]0[/sub])

where g = 9.8 m/sec[sup]2[/sup] is the acceleration due to gravity, u is the muzzle velocity of the bullets (in m/s), q is the rate at which mass is ejected (in kg/s), and m[sub]0[/sub] is the initial mass of the guy, his guns, **and** the bullets he has with him when he steps off the building.

Now suppose we want the guy to be able to step off the building, fire his guns straight downwards, and reach the ground with exactly zero velocity after some amount of time T. This means that the total mass of the bullets he’ll be firing is q T, and the mass of him and his guns is M = m[sub]0[/sub] - q T. If we want v(T) = 0, then we can jigger around the equation above to find that

v(T) = 0 = - g T + u ln (1 + q T / M)

or g T = u ln(1 + q T / M). If we plot the two curves y = g t and y = u ln(1 + q t / M), we find that they intersect at a non-zero value of T if and only if q u / M > g, or alternately, M < q u / g.

So what are the numbers? If the guns can really fire 800 rounds/minute, at 5 gms/round, and the guy has two guns, then q = 0.133 kg/s; u = 3100 ft / s = 940 m/s; and so our condition is that M < 12.8 kg, or 28.4 lbs. So unless the guy is **really** skinny, it won’t work (and don’t forget that this weight limit includes the mass of the rifles.)