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for physics/math people: how high would a bullet go if shot in the air? (1 Viewer)

Dennis Nicholls

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The air friction on bullets is extreme as they leave the muzzle in excess of the speed of sound. That, by the way, is why guns make that loud "bang" - it's a small scale sonic boom.

It's late so my mind is slow, but IIRC the speed of sound at sea level is around 1100 fps. Consider that the cheap Turkish 8mm ammo I shoot for grins has a muzzle velocity of around 2800-2900 fps and you see it's well over Mach 2 leaving the muzzle. I doubt it would ever return at over Mach 1 if I shot straight up with one of these Turk loads.

Think of it this way. Meteors hit the atmosphere at thousands of miles per hour and de-accelerate to only hundreds of miles per hour by the friction. At least what's left of them. Bullets will come back to Earth at a fraction of their cartridge's muzzle velocity.
 

BrianW

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I'm pretty sure this isn't true. Bullets retain their orientation gyroscopically as a result of their spin. Veering off course if allowed to tumble is how bullets follow the path of "least resistance."
 

Jeff Gatie

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Uhhh, I thought a gun makes a loud "bang" because of the explosion of smokeless powder inside the cartridge being expelled at great force out of the barrel, not because of the sonic boom from a small bullet, but I may be wrong. I guess all the subsonic .45 ACP ammo (~650-850 fps) I've shot is really silent and I just imagined the "bang".:D
 

VinhT

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I think there may be two noise components, a boom and a crack. :D For example, a suppressed rifle may reduce the boom factor, but does not mitigate the supersonic crack.
 

Jeff Gatie

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You are correct, of course. I just did not want anyone to think the only component of a gun's "bang" is a sonic boom. There is that little thing called an explosion going on also.;)
 

PhillJones

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Looks like you're right. I didn't realize hand guns had riffling, so I assumed it was a minimum resistance thing. It would appear that bullets are going too fast for that effect. It's obvious now I think about it since they are going supersonic so don't actually have streamlines.

If I'd paid attention in fluid dynamics I could probably tell you all about the sub-sonic / super-sonic transition. All I remember is that the physics of sub-sonic flight break down at the speed of sound. For this reason, people originally thought that the sound barrier couldn't be crossed.

Does anybody know what the drag co-efficient of a bullet is? the formula for terminal velocity was but a quick google search away. It's just are-arrangement of the drag co-efficients with the condition that acceleration = drag at Vt

Vt = sqrt ( (2 * W) / (Cd * r * A) )

w=weight, Cd=drag coefficient, r = air density, A = x-sectional area of object. Of course, picking a single X-sectional area assumes that the bullet is either a spehere or doesn't tumble but hey, it's better than nothing.
 

BrianW

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I slept through fluid dynamics myself, but I made up for it by watching Columbo.Wait a sec... Is this the TP thread?
 

Todd Hochard

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Just figure 200-250mph terminal velocity for a bullet that hasn't struck anything. That'll put a hole in your head.;)
 

reiling32

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So, how far would a 357 handgun bullet go if shot sraight across an open field before it stoped? Serious question!
 

Steve_Tk

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An open field? Is a bear crapping in the woods next to it? What happens if a tree falls in the woods next to the open field and no one is there to hear it, does it make a sound?
 

Jeff Gatie

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Given the distance from the ground when fired horizontally, the coefficient of friction is negligible in the Y direction, so the equation for time of flight would be D (distance above ground) = V1 (initial velocity) + (1/2)g (Earth gravitational constant) * t^2. Solve for t, giving you the time it takes for the bullet to fall to the ground:

D = V1 + (1/2)g * t^2

So say D = 1.5m (around 5 ft). V1 = 0, g = 9.8 m/s^2
1.5 = 0 + (1.2)g * t^2
1.5 = (1/2)*( 9.8) * t^2
1.5 = 4.9 * t^2
t^2 = 1.5/4.9 = 0.31
t = SQRT(0.31) = 0.55 sec Time of Travel

In a vacuum, the distance would be 0.55 sec * V1x (initial velocity in the X direction).
In the real world, the distance traveled is 0.55s * V1x minus the decelleration by the force of friction as described above. Since I don't have ballistics for the .357 handy, I don't have the initial muzzle velocity, nor the coefficient of friction of the particular bullet when in flight. So I guess that's as far as we can get.
 

DaveF

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Curious, I googled its ballistics.
http://en.wikipedia.org/wiki/.357_Magnum

Speed ~400 m/s, so about 220 m, or 240 yds (2.4 football field lengths).

(also, not nearly far enough that Earth curvature needs to be considered, which I was wondering about)
 

Joseph DeMartino

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Quote:
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 fact, under those theoretical circumstances, the bullet would land back in the barrel of the gun that fired it.

There have been a number of cases where stray bullets have killed people, sometimes people a considerable distance away. Some of these have even been bullets fired "in the air", but I'm not sure that there's ever been a case of a deadly bullet that was fired straight up into the air. Bullets fired at an angle (even a slight angle) retain much more of their initial impetus and, after reaching apogee of the ballistic arc, begin to accelerating again. (I had a distant cousin by marriage who died this way. She keeled over while cleaning her shower stall one day, in a closed, windowless bathroom. Her husband thought it was a heart attack until he saw a small spot of blood on her dress. The autopsy revealed that a bullet had hit her in the heart. They eventually traced the bullet back so someone target shooting blocks away who had missed his target.)

Mythbusters did one on the bullet question as well (and I thought their methodology was sound, in this case, which I don't always do.) They verfied that in most cases a bullet fired STRAIGHT up will reach terminal velocity and land with less-than-lethal forces. (IIRC they tested a 9 mm and a rifle round - either a thirty aught-six or a military round like a 7.62mm NATO round or a 5.56mm M-16/AR-16 bullet.) But they also consulted an emergency room doctor who verified that in rare cases serious wounds and even deaths were caused by bullets fire "in the air".

I think the answer to the OP's question is "No, a bullet fired straight up into the air will not return at its initial muzzle velocity, and will not strike with a fatal impact, but that SOME bullets fire 'in the air' at an angle can inflict fatal wounds."

Either way, shooting bullets up into the air is a pretty stupid thing to do.

Regards,

Joe
 

BrianW

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In fact, under those theoretical circumstances (perfect vacuum, no force but gravity), the bullet would land back in the barrel of the gun that fired it.
I must disagree, at least to the extent that such an experiment is performed on a rotating body such as the Earth (or moon, or some other vacuuminous (heh) body). The closer to the equator you are, the further toward the anti-spinward direction the bullet will land. On the North or South poles, however, your statement would be correct.
 

Joseph DeMartino

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I must disagree, at least to the extent that such an experiment is performed on a rotating body such as the Earth (or moon, or some other vacuuminous (heh) body).
But I specified that I was talking about theoretical circumstances in which there was a perfect vacuum and no other forces acting on the bullet. All of this would be contingent on shooting the bullet absolutely straight up, of course.

Later,

Joe
 

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