# How fast would a bullet travel through liquid air?

According to Wikipedia, a bullet from a .22 Long Rifle travels at a speed of 320 m/s and the density of liquid air is 870 kg/m3.

What equation would calculate the speed of a bullet fired into liquid air at a certain distance, say 10 meters?

The reason for the rather odd question is to see if a potential answer to a brain teaser is feasible.

Here's the teaser:

Alan fires a bullet from his gun and his friend Wade catches the bullet with his bare hands. The bullet does not touch anything but air after it leaves the gun and until it reaches Wade’s hand. Wade is uninjured. How does he do it?

• Well, if you're looking at answers in the realm of unlikely things like liquid air, there are other physical possibilities, like Wade is very fast, and grips the bullet from its sides at the bullet's initial speed and applies friction to slow it down, slowing down with it... or the bullet was fired straight up in the air and he's hovering in a helicopter in the upper atmosphere where the bullet finally stops before turning around... – YungHummmma Sep 17 '14 at 17:28
• I'm open to anything. Liquid air was the best I'd heard. I like the idea of Alan firing directly up. I'll have to calculate how high Wade would have to be in order to grab the bullet at 0 m/s. – Dan H Sep 17 '14 at 17:40
• That may be the answer. A .25 ACP (automatic colt pistol) would reach about 2,200 feet and the Burj Khalifa is 2,722 ft tall. If Wade was a window cleaner... – Dan H Sep 17 '14 at 17:47
• Wade and Allen are on rocket sled trains travelling in opposite directions? – user121330 Sep 17 '14 at 17:52
• I agree with Dan H. Wade can be moving as fast as the bullet and catch it from sideways – user59317 Sep 17 '14 at 17:54

The speed of the bullet is not a constant. The value given of 320 m/s most likely refers to the barrel exit velocity of the bullet. For a quick and dirty approximation to how the bullet speed varies with distance you can simply use Newton's famous equation $$F = ma$$ With $F$ being the total force acting on the bullet, $m$ the bullet mass, and $a$ the acceleration. In this case, the force acting on the bullet is primarily due to drag from the fluid. Assuming supersonic flow, to accurately calculate the drag you would need to know something about the shape and equation of state of the fluid. From there you could estimate the flow over the bullet and come up with an approximation to the force exerted on the bullet by the fluid. However, how to do that is probably too much detail for the answer that you are looking for. So, lets assume that the drag force acting on the bullet is given by $$F = \frac{1}{2}C_d \rho u^2A$$ Where $C_d$ is the drag coefficient measured from experimental data, $\rho$ is the fluid density, $u$ is the bullet's speed and $A$ is the frontal area of the bullet in the direction of travel. Unfortunately, in reality, $C_d$ is not a constant value across changes in density and velocity. However, lets assume that it is, then we get $$-\frac{1}{2}C_d \rho u^2A = ma$$ or with $$a = \frac{du}{dt}$$ we get the system of ordinary differential equations for position and speed $(x,u)$ $$-\frac{1}{2}C_d \rho u^2A = m\frac{du}{dt}$$ with $x$ the position of the bullet $$u = \frac{dx}{dt}$$ Or, since we are interested in the speed as a function of distance, using the fact that $$u\frac{du}{dx} = a$$ we have $$mu\frac{du}{dx} = -\frac{1}{2}C_d \rho u^2A$$ giving $$\frac{m}{u}\frac{du}{dx} = -\frac{1}{2}C_d \rho A$$ Integrating gives $$Log(\frac{u}{u_0})= -\frac{C_d \rho A }{2m} x$$ or $$u(x) = u_0 e^{-\frac{C_d \rho A }{2m} x}$$ From that you can estimate how much difference a change in density makes. A very rough estimate, but an estimate none the less.