How fast does a bullet return to earth?

  1. My question is based on a stray bullet that was shot on new years, and we discovered the it had hit the surface of our tennis court with such force that it made a hole aprox. the same size as the bullet it's self. A group of friends discussed different opinions as to how fast the bullet comes down oposed to the speed it leaves the gun, or said in other words, does it come down faster than it went up?

    If anyone has a general answer (taking into consideration we do not have the exact launch speed, wind, bullet weight, etc.)


  2. jcsd
  3. ahrkron

    ahrkron 731
    Staff Emeritus
    Gold Member

    In a perfect vacuum, the bullet would return (to the height from which it was shot) with the same speed it had when it left the gun's barrel; in an atmosphere, part of its energy will be lost in collisions with gas molecules.
  4. it would take the same amount of time for it to go back as it would coming down. since gravity is constant [close to the surface of earth for all practical purposes, anyways] it'll act upon the bullet with the same amount of 'pull' [for lack of better word :P]

    its terminal velocity depends on on its size and mass, but i doubt it would reach. bullets are pretty aerodynamic tho, so they do go pretty fast. but you could expect it would land with roughly the same speed as it left.
  5. I disagree: the speed of a bullet leaving the gun is typically (much) faster than the speed of sound: ~1000 m/s. Compare this to the terminal velocity of a skydiver: ~100 m/s. I'd say that there is no way that a free-falling bullet would break the sound-barrier (~300 m/s), so somewhere arround ~100-200 m/s seems like a reasonable estimate of a free-falling piece of metal.
  6. I agree, it definitely does not come down as fast as it went up. If it is shot from horizontal in a vacuum, it will reach the ground at the same time as another body dropped at the height of the gun. In air, the dropped body would hit the ground first because the fired bullet is slowed down more.
    There are equations for particle paths for air resistance that include atmospheric density and the velocity of the particle etc, but they are large.
  7. Integral

    Integral 7,287
    Staff Emeritus
    Science Advisor
    Gold Member

    This is a question about the terminal velocity of a bullet, a dense streamlined object designed to travel at high speeds through the air. Why would we base estimates of this on the terminal velocity of the human body? While I do not know what the terminal velocity of a bullet is, I am sure equal to or 2 twice that of a human body is on the very low end of the possibilities.
    Here is a good link.

    Actually the terminal velocity of a human is more like 120MPH this is about 54 m/s while that of a bullet seems to be anywhere from 100~400m/s depnending on the size and mass of the slug. (Note the high end is for a 14" Naval gun!)
    Last edited: Feb 16, 2004
  8. It is rather obvious that the bullet will land at a lower velocity - simple conservation of energy, and second law of thermodynamics.
  9. 400 m/s would mean that the free-falling bullet breaks the sound-barrier. Do you have a link proving this (I find it hard to believe)? I could not find this info in the link you provided.
  10. Integral

    Integral 7,287
    Staff Emeritus
    Science Advisor
    Gold Member

    That is from the last paragraph of the link, calling a 1/2 ton of steel a bullet is a stretch. That is called a shell, Looks like the reference is given.
  11. Regarding the possibility of the bullet breaking the sound barrier in its fall - it is possible for objects to break the sound barrier while falling - just ask Joe Kittinger (google if you don't know who he is). However, of course, he exceeded the sound barrier at a very high height, where the air was thin.

    In order to see whether the bullet could exceed the sound barrier, lets look at the energy gained and lost per distance fallen. The bullet will gain energy +E=d*g*m, and lose energy -E=d*Cd*p*v^2*A/2.

    d=distance fallen
    m=mass of bullet
    Cd=drag coefficient
    p=density of air
    v=velocity while falling
    A=frontal area (from which Cd is generated)

    Setting -E=+E, and solving for v, we find that v=((2*g*m)/(Cd*A*p))^1/2.
    Cd at transonic speeds is about 0.45 for a bullet with an A of 5.8x10^-5 m^2 (a 7.62mm rifle bullet, AK-47). p at sea level is about 1.275 kg/m^3. m for a 7.62 mm bullet is about 0.01 kg. The terminal velocity calculated by this equation is about 76 m/s, well below transonic speed. Using the subsonic drag coefficient of 0.3, we get a terminal velocity of 94 m/s. So the bullet would come down at nowhere close to transonic speed. Nevertheless, a bullet hitting you at 94 m/s has a kinetic energy of 88 Joules, spread over a small area. It won't hurt as much as a regular bullet shot, but it will still hurt you. I'd wear a helmet if I were firing bullets in the air - people die all the time from "celebratory firing".
  12. Integral

    Integral 7,287
    Staff Emeritus
    Science Advisor
    Gold Member

    Perhaps you did not notice..

    The shell that may be breaking (is breaking according to analysis similar to yours) is from a 12" navel gun. m= ~2.2x103kg
  13. in my opinion the bullet shot up would be faster than the bullet dropping by the gravity force. when shooting up the potential energy should be greater than the potential energy of the bullet falling. even though when falling the gravity acts on it.
  14. A bullet is said to travel 33,000 MpH which would obviously lose it's velocity at an upward angel. Depending on the angel in which it was shot it, it's hard to tell exactly how fast it made contact with your tennis court. To determine the speed of the bullet would be a large process involving you figuring out the approximate angel of entry, the type of bullet, gun, atmospheric pressure and other weather factors aswell as where the person shot the gun from. You would probably need to make some sort of quadratic sketch to find this out. Sorry I couldn't actually answer the question however, this should help you if you so choose to go to these great lengths.
  15. ZapperZ

    ZapperZ 30,559
    Staff Emeritus
    Science Advisor
    Education Advisor

    Wow! Talk about digging up the dead! Did you know that you're responding to a post from 2004?

  16. russ_watters

    Staff: Mentor

  17. Mythbusters are made-for-tv idiots. Look at their pathetic excuse for physical reasoning:

  18. russ_watters

    Staff: Mentor

    Read a little further. They constructed a vertical wind tunnel and dropped bullets into it.
  19. I don't agree. The horizontal velocity of the fired bullet is irrelevant in this instance.

    Both the "stationary" object dropped and the bullet fired begin with zero vertical velocity. They will accelerate downwards at the same rate, therefore hit the ground at the same time.
  20. I note this thread started some two years ago, however it is an interesting question. Friction due to air resistance is a very tricky thing according to my research. A great many factors will affect this value.

    In the original post, there was not suffcient 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:

    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 300g, drag coefficient 0.228. The "nose down" attitude gives us a cross-sectional area of 0.0001026m^2.
    4. Still day.
    5. Ignoring humidty.

    We can use the Quadratic drag formula.

    [tex]v_{terminal} = \sqrt{\frac{2mg}{CP_{air}A}} [/tex]

    As I right this, LaTeX isn't working, so that is saying terminal velocity =
    square root ((2mg)/(CPA)).
    m = mass
    g = acceleration (gravity)
    C = drag coefficient
    P = air pressure
    A = cross-sectional area

    That comes out to about 442m/s = 1,448ft/s = 987mph.

    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 0.0002052m^2 Plugging those numbers in gives:

    192m/s = 630ft/s = 430mph.

    That should be good enough as a "useless information" tidbit at your next BBQ.
    Last edited: Aug 21, 2006
  21. This is interesting, from "Hatcher's Notebook":

    "Out of more than 500 shots fired after adjusting the gun--only four shots hit the platform. One of the shots was a service 30.06, 150 grain flat based bullet, which came down base left a mark about 1/16 inch deep in the soft pine board...

    It was concluded from these tests that the return velocity was about 300 feet per second. With the 150 grain bullet, this corresponds to an energy of 30 foot pounds. Previously, the army had decided that on the average, an energy of 60 foot pounds is required to produce a disabling wound. Thus, service bullets returning from extreme heights cannot be considered lethal by this standard...

    This supports the observations of those who wrote during WW2, that after a heavy battle, a number of bullets were found slightly embedded in tar rooftops, all pointed towards the sky.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?