At what speed would a wind affect a bullet? Firing a gun loaded with the fastest bullet (.220 Swift 1,422m/s or any bullet that is super fast and excellent aero dynamics) in a close range (2cm) from the tip of an air blower.  What would be the speed of the air coming out of the air blower to be able to deflect the bullet off course 90 degrees?
 A: For the bullet traveling directly at the air blower, to stop the bullet within $2cm$ needs, from the equations of motion a deceleration of $5\times10^{7}m/s^{2}$
Air resistance is $$F=\frac{1}{2}\rho ACv^2$$ see for example this website
For air $\rho = 1.2$, $C=0.2$ (estimate) and $A=\pi r^2$ with $r=2.8\times 10^{-3}m$
so from $F=ma$ with a mass of the bullet of $2g$
$$0.377r^2v^2 = 2\times 10^{-3}\times5\times10^{7}$$
$$v^2 = 3.4\times10^{10}$$
$$v = 184,000m/s$$
We could subtract the speed of the bullet from this, but it doesn't make much difference.
A: Remember that the bullet does not know what the wind speed is.  The bullet only knows to travel in its given medium.  So if the wind was blowing from the side at the same 1,422m/s, then the bullet would travel sideways in that medium at the same rate that it is travelling forwards.  In this case, in one second it would travel forward 1,422 metres and sideways 1,422 metres, so it would travel at a 45 degree angle. This might be the answer you are looking for.  About 5,000 kph.
Otherwise, to travel in a complete 90 degree angle, it would have to have no forward travel at all, and complete sideways travel from the moment it leaves the gun.  So if you said that in the first millisecond after leaving the gun, it travelled directly sideways, then the wind speed would have to be 1,000 x 1,422 = 1,422,000 metres per second.  Darned fast.
But in general, any wind speed at all changes the path of a bullet.  This is a key aspect of being a target shooter, and is the reason why they set up ribbons along the rifle range, to show the wind speed at different points.

