If a bullet is fired straight up in the air, will it return to ground level with the same velocity?
Some labs are not safe to perform in schools, but give valuable insight on physics: Thus we may watch a video of such an experiment.
MythBusters Episode 50 – “Bullets Fired Up” – Busted, Plausible, and Confirmed
All guns are “rifled” in order to impart spin to a bullet leaving the barrel.
This rifling gives the bullet angular momentum, which stabilizes it (keeps it from tumbling)
How can we see a bullet’s angular momentum?
Rhett Allain offers a physics analysis of the “Bullets Fired Up” episode of MythBusters
Results are complicated: In most cases the bullet lands with non-lethal velocities, despite confirmed cases to the contrary. How to solve this apparent paradox?
There are really two different cases:
Experiments show that bullets lose (*) spin (and angular momentum) via interactions with the air. Once they lose angular momentum they are no longer spin-stabilized. Without spin-stabilization they tumble, and thus they experience more drag (“air resistance”), slowing them down more.
Bullets fired at an angle can come back down fast enough to kill someone. Experiments show that this allows bullets to keep more of their spin/angular momentum. They stay spin-stabilized for more of the trajectory, and thus keep more of their energy upon landing.
(*) Both momentum and angular momentum are conserved quantities; when we say that an objects loses either, it is because the momentum is actually transferred at a microscopic level to the surrounding molecules of air. See our lesson on momentum.
Analysis: What if the bullet falls without tumbling? How fast would they go
PhysicsForum meber WhyIsItSo gives us the following:
In the original post, there was not sufficient information to say, for example, if the bullet still had kinetic energy from being fired, or if all the speed that was left was its terminal velocity. For the latter case – Assumptions:
1. Take the ideal case (unlikely) that the bullet is falling in a consistently “nose-down” attitude.
2. Air pressure is 1.29Kg/m^3
3. Use a .45 Cal bullet, mass 300 g, drag coefficient 0.228. The “nose down” attitude gives us a cross-sectional area of 0.0001026m2.
4. Still day [no winds].
5. Ignoring humidty.
Process: We can use the Quadratic drag formula.
m = mass of bullet
g = acceleration of gravity = 9.8 m/s2
C = drag coefficient
ρ = air pressure [This symbol is the Greek letter rho]
A = cross-sectional area of bullet
That comes out to about :
Velocity terminal = 442 m/s = 1,448 ft/s = 987 miles/hour.
Gun enthusiasts are probably thinking that number is higher than the muzzle velocity of most guns, which means that in this “ideal” scenario, a bullet fired straight up wouldn’t get high enough to attain terminal velocity on the way down.
Much more realistic would be to assume the bullet is more or less sideways on the way down. This will result in a drag coefficient more like 0.6 (a sphere is about 0.5), and a cross-sectional area – very roughly – twice the nose-on area, or say 0.0002052 m2
Plugging those numbers in gives:
Velocity terminal = 192 m/s = 630 ft/s = 430 miles/hour