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

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.

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?

Impposible to tell without the following:

- Muzzle velocity of the bullet
- Drag on the bullet as a function of it's shape




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

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.

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.

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

At the very least, we would need to know the initial velocity of the bullet. Actually that's really all you need. Everything else can be approximated or neglected. Mass of the bullet is irrelevant if you know the initial velocity.

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

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).

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.

1- what the heck is terminal velocity?

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

--
H

werd.
K.I.S.S.

Terminal velocity is the maximum velocity that can be aquired through freefall in the earths atmosphere (or any drag inducing medium). Since the coefficient of drag is a function of velocity, the force of drag increases as the velocity of an object increases. Eventually, the force of drag overcomes the gravitational force, thus ceasing the accelleration of the object, causing it to fall at a stable, constant velocity.



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

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

Interesting, I did not know that. Yet it makes perfect sense, thanks.
What Phillip said .

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H

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.

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.

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).

man, early physics classes must have just irritated the shit out of you.

mythbusters irritates me. last week I saw a show testing the difference between driving with your windows down vs. A/C on, they used TWO DIFFERENT VEHICLES!?? WTF? No two are alike, period.


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

The bullet will reach one to three miles before falling back to earth.

Basic equation for motion:
D(t) = d + v*t + 1/2*a*t^2

where
t = time (sec)

do = initial height = 0 m

v = velocity of bullet out of gun = 300 m/s (~984 ft/s, based on Philip's info)

a = gravity = -10 (-9.81 m/s^2 really) (gravity)

d = distance as a function of time

Wind resistance ignored.
Motion assumed to be perfectly vertical.

Employ some calculus and solve for maximum distance:
D'(t) = d/dt D(t) = v + a*t
D'(t) = 0 = v + a*t (at max)
v = -a*t (at max)
tmax = -v/a = - 300 / (-10) = 30 sec

Max distance:
D(tmax) = 0 + 300*30 - 1/2*10*30^2 = 4500 m = 4.5 km = ~ 2 miles

Get a freshman or sophomore engineering student in here to check my math. I've not done this for homework in over a decade

No, I liked them. I just knew they were not applicable to the real world, hence I became an engineer. My HS physics teacher used to have a thing called the "bump factor", where each kinematics experiment was given a small "bump" to factor in all the stuff we ignored. Got the ball in the cup (or whatever the "successful" experiment was) every time after the "bump factor" was applied.

Phill - you snuck in your answer while I was typing mine!

As others have correctly stated, the muzzle velocity only matters if the gun was fired in a complete vacuum and the bullet was allowed to fall without any outside forces besides gravity, in which case the bullet would land with the same velocity at which it was fired.

In practice, it's all about terminal velocity. For example, in theory a human falling should continue to accelerate as they fell, but in practice they max out at around 120 MPH (my number here may be off by 40 MPH or so, but that's about what I remember).

Likewise, the Mythbusters tested to see if a penny dropped from the empire state building would actually kill somebody, and they measured that the terminal velocity of a penny is between 30 and 60 MPH (it has much less mass and it's shape causes it to tumble and interact with the air much more than a falling human would). They determined that a 60 MPH penny would not cause serious harm to a person.

When a bullet leaves the gun, the rifling makes it spin and keeps it streamlined against the air. I suspect that when it reaches 0, and then starts to fall, it would also start to tumble and interact with the air a great deal (although much less so than the penny). I have no idea whether such a bullet would be lethal or not, but I suspect that it would be heavily dependant on it's mass.

physics is fun, math isn't

All of that math is giving me a headache!

Your other question is yes, people have been killed from falling bullets.

I can't remember where/when exactly, but it has been on the news a few times. Some cultures celebrate by firing guns into the air - a truely insane idea.

I keep on thinking Afghanstan after they were freed, but weedings and New Year's Eve also comes to mind.

I've seen people do the arrow thing - there would have to be ZERO wind, and the arrow would be tumbling when it changed direction (from up to down), so having it land in the same spot is unlikely.

Glenn

The bullet from a gun doesn't stay pointing forward just becuase it was in that orientation in the barrel. It does so because it's the orientation of least resistance. It doesn't really matter, which orientation the bullet starts off in, it'll align itself very quickly.

I would imagine a coin is a bad shape for drag, the sharp leading edges would liekly set up vortices at the side and make for poor air flow.

off topic..

I liked that episode. If what you are saying is your main complaint, then why bother with the experiment? Even if they used the same automobile, you could argue that it would yield different results with a different car. Plus No two drives are the same either. You can argue that when the first drive ran out of gas, it messed up the engine for the second drive.[/rant]

Back on topic, this is what google has to say about the origianl question:
http://www.villman.com/forums/topic.asp?TOPIC_ID=1087

It talks about a couple people that have tried it too and I found the results interesting.

Somebody got killed by a bullet fired into the air in an episode of Homicide: Life on the Street, which is a pretty realistic show.