# for physics/math people: how high would a bullet go if shot in the air?

Discussion in 'After Hours Lounge (Off Topic)' started by NickSo, Jun 14, 2005.

1. ### NickSo Producer

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a friend asked me this just a few minutes ago...

well, actually here's how the convo went:

now im just curious... out of a basic handgun (dont know which, i guess whichever is the most common), if one were to shoot a round straight up into the air, how high would it travel?

and once it stopped going up, and started going down, how fast would it be going when it reaches the ground?

2. ### Philip_G Producer

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if shot straight up it will decel. at 9.8 m/s/s from the muzzle velocity, to zero. So it would depend on the muzzle velocity.
Ditto on the way down, accelerating at 9.8 m/s/s until it hits something. Since the distance up and down are the same it will hit at muzzle velocity, negating friction and other imperfections.
I'm sure one of the engineer types can giv ea better explanation, but that's the basic physics 101 of it.

3. ### Ted Lee Lead Actor

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so in english philip...?

sounds like you're saying, barring friction, etc -- a bullet comes down about as fast as it left the chamber? if that's the case, then it certainly could kill someone.

in a lot of places, isn't it illegal to shoot a gun in the air for that exact reason?

4. ### Jeff Gatie Lead Actor

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It would (eventually) reach terminal velocity (Vt). Again. we would have to know the mass and shape of the bullet, plus how far it was from the ground, but the equation would be: Vt = {2*m*g/[Cd*A*r]}1/2

5. ### alan halvorson Cinematographer

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I can't answer your question, other than to say that it would depend on what sort of bullet was being shot, and that regardless of the bullet, its speed on the way down would be substantially less than going up.

I calculated many years ago that a SR-71 Blackbird spy plane at its cruising speed was faster than a bullet shot from my Ruger 7mm deer rifle. Useless but interesting trivia.

Many years ago, when I was into bow hunting, I stood in an open field and shot an arrow as straight up as I could, which turned out to be not that straight. I was not thinking too clearly at that moment. As soon as I released the arrow, I realized that if I really did shoot straight up, I was in trouble and no matter how straight I shot, other people could be in jeopardy. Fortunately it came down about 100 yards from me. Taught me a lesson.

6. ### Philip_G Producer

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in a perfect world the accel and decel would be the same, so... yep. Not the chamber but the barrel though.
Jeff's explanation is better and more real world.

if you really wanted to figure it out most of my loading manuals give you the coefficients for different bullet weights and calibers, you could calculate it with a hypothetical but accurate muzzle velocity.

a fun demonstration I remember from physics class involved a long tube and a ball bearing used as a blow dart, and a metal can hanging from a magnet, on the end of the tube was a switch to cut power to the magnet as the ball bearing left the "barrel", the bearing and the can would hit in mid air since both started falling at the same rate/same time.

7. ### NickSo Producer

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what kind of gun/bullets do the police use? maybe use those specs?

8. ### Philip_G Producer

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depends. Some 9mm some 40 smith, some 45acp, some 38, some 357

here's a 40 smith.
we'll use a 150gr nosler bullet, ballistic coefficient is .106 and the sectional density is .134, muzzle velocity will be best case about 1170fps.

someone else can hack the math, you've gone beyond my physics "expirtise" now

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Obviously that depends on how high it went in the first place...

If you don't wanna go through all that, I am sure there is empirical data somewhere online.

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H

10. ### Jeff Gatie Lead Actor

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Again, that's hard to say. Anywhere from a .38sp to .357 to 9mm to .40S&W to .45 ACP are in use today (plus many more), sometimes within the same police force. Most popular for inner city are the 9mm and .40S&W, but each load and bullet are very different. The difference between say a 9mm (light bullet, high velocity) and a .45 ACP (big, heavy bullet, real slow velocity) would be so large as to make the exercise moot, if applying it to "bullets" in general. We really can only do specific calculation, most of which (computing drag) are diffcult to impossible for the layman (without a ballistics chart handy).

11. ### Philip_G Producer

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within the context of the question the data I've provided should be more than accurate enough. Depending on load and powder the variation of velocities was less than 200 fps for a 150 gr. load.

if you wanted to be more accurate you could look up the data for .40 S&W federal hydrashoks, I would guestimate that 90+% that are shooting 40 smith are shooting hydrashoks.

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1- what the heck is terminal velocity?

2- Mass is irrelevant in a free fall, exercise, which is really what this is.

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H

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werd.
K.I.S.S.

14. ### Jeff Gatie Lead Actor

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Actually, it was a real world question. You know, the real world, that place where us engineers live.

15. ### Philip_G Producer

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I figured the esteemed Mr. ¿§êrìôù§«£òvëñ»? would do with a basic answer

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What Phillip said .

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H

17. ### Jeff Gatie Lead Actor

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But that would not be a correct answer. It's the old "drop a penny from the Empire State building and it will be like shooting someone in the head" myth. Terminal velocity will be reached by the penny long before it reaches a deadly velocity (see the episode of Mythbusters where they did just this). I suggest a bullet would do the same, given the wider cross section.

18. ### PhillJones Second Unit

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Anybody call for a physicist.

Neglecting friction and assuming that you fire it directly up in the air, it's a dead simple question.

u=initial velocity
v=final velocity
a=acceleration
s=displacement (distance from origin, ie height)
t=time

v^2=u^2+2as (I could derive this but can't be bothered to figure it out)
now the intial velocity is the muzzle velocity, the final velocity is zero (we're interested in when the bullet stops and starts to fall back down), the acceleration is -9.8. I'm choosing upwards to be the positive direction, therefore gravity is negative.

0=u^2 - 20*s
s=u^2/20

So the height is the muzzle veoloctiy squared, divided by 20.

The time it took would be:
from v=u+at
t=u/10
the muzzel velocity / 10
The time to hit the ground is twice that, by symmetry.

You can calculate the velocity of the bullet at any time using this equation
v=u+at
or any height using s=ut+(1/2)*a*t^2.

You have to make sure that you stick to your sign convention that velocities upwards are positive and those downwards are negative.

If the bullet wasn't going straight up, you treat the horizontal and vertical components of it's velocity seperately.

The vertical component would be
u_v=u*cos(a) where a is the angle from straight up.

The horizontal would be
u_h=u*sin(a)

The horizontal velocity is constant so once youve subbed the result for u_v into the equation for time to hit the groud, you can calculate how far the bullet travelled horizontally just by using velocity=distance/time.

19. ### Jeff Gatie Lead Actor

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Jeez, we don't need a physicist for that. I took those classes in High School.

The answer is, if we lived in a vacuum, what both Phil(l)'s said. Tell you friend not to fire bullets in a vacuum and he's all set (ohhh, the irony).

20. ### Philip_G Producer

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I do find it kind of frightening that neither of them could come up with the basic explanation.