How far will a bullet travel if it does not hit anything?
98210
depends on the gun/cartridge type and how hot it is loaded\nsome well over a mile
882491
#8 pellets will damage out to a hundred and 36 feet provided the temperature is around 50 degrees or more.  At temperatures of less than 50 degrees the pellet will travel farther due to the air being less dense.  There is a proven factor--try Federal Cartridge Company or DuPont's web pages for more precise data.
334471
If fired straight up, no the terminal velocity is too small.  But if fired at an angle, the bullet maintains horizontal speed and that can be enough to kill.
510542
16.5 to 17 yards. It all depends on how far it needs to go. The further it needs to go the closer it is to 17 yards for trajectory reasons.
892312
Hmm-m. Interesting question...\n\nI thought about this for a while and see no\nreason why this math would not be correct.\nAssuming that the bullet did explode when\nstruck by the firing pin, and the bullet left \nthe end of the barrel of the gun at say \n2000 feet per second velocity...\n\n2000 fps = 1800 Mph\n\nThe Moon is 239,000 miles from Earth.\nLet's say you were 10 miles up in the sky\nwhen you fired the gun.  Distance = 238,990 Miles.\n\nSo, 238,990 / 1800 = 132.7 hours \n\nSince there is no atmosphere in space to slow down the \nbullet as it traveled, we can estimate that the bullet would \ncontinue traveling at 1800 mph for the entire trip. The \ngravitational pull of the Moon would influence the bullet's\nvelocity somewhat towards the end of its flight, but my\nfigures show that with 1/6 the gravity of Earth, the Moon\nwould not significantly change the time of flight.\n\nAnswer: 132 Hours.   or   5.5 Days.\n\nNow, I have two questions for you:\n\n#1.)  If the rifle did fire off in space, would you see any smoke?\n\n#2.)  If you were 10 miles up in the atmosphere - say 50,000 ft.,\nhow fast must you be going, to be able to remain at that altitude,\nand in which direction?\n\nYour question makes reference to the Space Station as the point of firing the gun. In my answer above I did not use that as the platform from which the gun was fired because I do not have at my finger tips the approximate altitude of the Space Station though I know it is much more than 50,000 Ft. I also did not take into account the thin atmosphere which exists at 50,000 feet, or the pull backwards on the bullet by the gravity of Earth. These omissions were made simply to run the numbers and see what I came up with since I thought this was an interesting question. Now, having dispensed with the mundane aspects of my quick math, let's discuss the real problem with this question...\n\nThe Earth is spinning. The Moon is orbiting the Earth at a certain velocity, hence it is a moving target. The moving target is being shot at from a moving platform. If you have any hope of hitting the Moon with your bullet, you would have to fire at the place where the Moon will be 5.5 days from the time of firing your gun, not at where the Moon is now. To do that requires more math than a few simple divisions of distance and velocity. For example, the moon swings completely around the Earth at a radius of 238,900 miles in 27.3 Earth days. The circumfrence of that circle is Pi x D = 1,500,292 miles (assuming the orbit is a circle, which it is not). So, the velocity that the Moon is moving away from its present spot right this minute is 2,290 miles per hour.  Since the Moon rises and sets on our horizon each night, you can see that in our calculation, the Moon would have passed overhead at least 5 times in 5.5 days. So your rifle would have to be pointed at the place you calculate the Moon should be at in 5.5 days.        \nRegards,\nZah
251965
The bullet would disintegrate in the Earth's atmosphere.\nExplode?  From a bullet?  Uh, no.
143519
So something small like shrapnel or a bullet wouldn't kill them before they got to where they were going.
37339
Assume the bullet is striking the door at a right angle to the door. This is not stated, a which is an odd thing to leave out of the question. I mean think about it, if the bullet hits the narrow side of the door edge on, the door will not turn at all. D'oh!\n\nSo anyway to answer the last part, over all energy is conserved, but kinetic energy is not conserved. Some of the bullet's kinetic energy is lost in damaging the door, and some is lost to frictional heating of the door, and some is lost in friction with the air. Hammer a nail and it gets bent and hot. But let's pretend that this is a moon door and there is no air, and that the energy lost in heating and deforming the door is too small to measure. \n\nThe Kinetic Energy (KEb) of the bullet is 1/2 x mass x velocity x velocity. Bullet's mass is 0.01 kilograms. Velocity is 400 meters/second. So KEb equals 1/2 x 0.01 kg x 400 m/s x 400 m/s. This is 800 (energy units).\n\nIn this rotation problem, instead of mass, we use "moment of inertia of the door" or "Idoor". And instead of velocity we use "angular speed" or "omega"\n\nFor rotation problems the Kinetic energy (of the door) equals 1/2 x "Idoor" x omega x omega\n\nWe can ignore that the door is heavier with the bullet stuck in it. And we can pretend that all the mass of the door is squooshed down into a small lump at a distance (radius) of 0.67 meters from the hinge. This is a really poor assumption, but since I can't guess the background of this problem (high school, engineering school, freshman physics), or whether the teacher wrote the problem or got it out of a book. Or if the teacher is a simplifier or a stickler, we will go with the poor assumption. You'll get points for trying anyway.\n\n"Idoor" equals mass x radius x radius. So "Idoor" is 10 kg x 0.67 meters x 0.67 meters. Equals 4.4 (rotational inertia units).\n\nSo pretending that all the bullets Kinetic Energy (KEb) is zapped in to the door's Kinetic Energy (KEdoor) we say KEbullet = KEdoor\n\nSubstituting using Idoor:\n\nKEbullet = 1/2 x Idoor x omega x omega\n\n800 = 1/2 x 4.4 x omega x omega\n\nsolving for omega:\n\nomega = square root of (800/2.2) = 19 radians per second (APPROXIMATE ANSWER)\n\nA second hand on a clock moves at 6.28 radians per 60 seconds, or about 0.1 radians per second. So the "moon" door is whipping around about 190 times faster than a second hand.\n\nIn electronics, a "jiffy" is 1/60 of a second. Imagine a clock with a jiffy hand. A jiffy hand zips around a clock face in one second. That is about 6 radians per second. So the door is moving about 3 times faster than a jiffy hand.\n\nGood luck! Remember with most science teachers the numerical answer is less important than the reasoning and steps you use.
1079082
bullets are usually mafe of lead, sometimes covered in steel for long-range accuracy.  Bullets normally do bounce (ricochet) off hard surfaces.
937015
It varies very widely - it all depends.\n\nSome handgun projectiles are as slow as 600 feet per second.\n\nSome rifle projectiles are as fast as 5000 feet per second.\n\n(Tell him much faster than a race car!)
30590
nothing's wrong with the above and your conversion is correct.\n\ndoesn't seem like very much eh?  the reason is b/c of what i explained in the other question.. if you plot a graph of the force versus time, it will look like this:\n\n_____|___|___|___|___|___|___|___|_____\n\nthose sharp vertical lines represent the bullet impacts, and the rest of the time the force is zero.  if you average over this time span, the number turns out to be very low.  but the height of those peaks is very high because so much momentum is exchanged in a very short time!  so force is kind of misleading.  in physics the more meaningful way to talk about collisions is to talk about the momentum exchanged in a collision, which is called an "impulse" because it happens over such a short time.\n\nput another way, the bullet has a momentum of m * v, or .003 kg * 740 m/s = 2.22 kg*m/s.  think about what that means.  it's like getting whacked in the chest with a 2.22 kilogram (around 4.5 lbs) object moving at 1 meter per second - pretty heavy and pretty fast!  \n\nand that's just one bullet.  think about 8 of them hitting you every second.  furthermore, 3 grams seems like pretty low estimate for the mass of a bullet.   that's something like a .22 or a .25.  consider that a .44 magnum bullet, which is the sort of gun that actually can knock you off your feet, has a mass of 250 "grains" which is closer to 16 grams.  that's more like getting hit with a 10-kg object.  that's like someone winding up and throwing a 22-lb weight at you!\n\nso there you go.  the averaging and the short time interval for the impulse is what's misleading you.
1012557
maximum effective range at a point target is 550 meters.  at an area target is 800 meters.  Maximum range is around 3500 meters (meaning you can hit something that far out, but it's a big crapshoot).
950828
Gravity only "slows" the bullet during 1/2 of its total trip, during the other half, gravity accelerates the bullet.\n\nAs the bullet rises into the air its velocity is continually decreasing as the force of its own weight (force due to gravity) as well as the force of air resistance, act to accelerate it downward.  At some point the bullet will come to a stop and then begin falling.  Afterward, gravity acts to increase the bullet's speed while the air resistance works in the opposite direction.  The magnitude of the force of air resistance depends on the speed of the bullet (the faster it does, the harder the air pushes back on it).  Eventually, the bullet will reach what is known as the "terminal velocity", where the force of gravity equals the force of air resistance, and the bullet stops accelerating (net force = 0).\nThe terminal velocity for an object depends on several factors including the object's mass and shape, as well as the density of the fluid it is moving through.  The terminal velocity for a bullet is lower than the speed at which it comes out of the gun in most cases, so the bullet will never be traveling as fast as it was while leaving the barrel when it hits the ground...but still, I would still not want it to hit me.
239661
The bullets won't be pulled to earth at the same downward _speed_, but they will be pulled by the same downward _force_.  The force that pulls it toward the earth (gravity) would cause both to have the same vertical speed, assuming:\n\n(1) the bullet that was shot from a gun is fired horizontally, and\n(2) these occur in a vaccuum, where the effects of air can be neglected.\n\nNow, to find how long it would take them to hit the ground and how far away the fired bullet would be, you'd need to know how high the bullets are initially and how fast the bullet came out of the gun.  There are a few formulas that you can use:\n\nvf = vi + at\nvf^2 = vi^2 + 2ad\nd = vt\n\nwhere\nvf = final velocity\nvi = initial velocity\na = acceleration due to gravity\nt = time\nd = distance\n\nUsing these in a logical manner can get you what you need.  For example, you can use the second equation to find the vertical velocity of the bullet as it hits the ground.  Then you can take this vertical velocity (vf now) and use the first equation to find how long it took to fall.  Then with this time, you can find out how far the fired bullet went using the third equation.\n\nIt helps if you separate the bullets' velocity into a vertical and horizontal component - i.e. how fast is it moving horizontally and how fast vertically.  These two are independent of each other.  If you keep that in mind that helps in understanding why, in a vaccuum anyway, these two bullets should hit the ground at the same time.
164638
I must bebunk the answerers who think the bullet will be travelling faster without the absence of gravity - the answer is yes and no - yes if compared to a bullet fired vertically on earth or at any angle, no if compared to bullet fired horizontal to the ground on earth.\n\nIf bullet is shot horizontally here on earth, and then shot in space, both bullets will travel at the same speed. The bullet travelling horizontally on earth will have gravity only affect its vertical movement, hence it will be pulled faster and faster towards the ground, but its horizontal velocity will stay the same.\n\nStraightdope has covered this in an article:
890955
the bullet travelled 86 cm in the time it took the disks to travel 0.6 rad.  So the time it took to travel that distance was 0.6 / 93.2 seconds.    The bullet speed is then 86/ <that number> cm/sec
726299
chemically, nothing happens to the bullet itself.  the gunpowder exploding is however a chemical reaction. \n\nMechanically, a pressure builds up behind the bullet and starts pushing it down the barrell.  Some bores are smooth (shotguns) and the bullet goes straight down the barrell.  Some barrels have ridges cut on the inside in a spiral pattern that grip the bullet and put a spin on it.  This increases distance and reduces the effects of wind.  The bullet continues to accelerate down the barrell to the muzzle.  At the muzzle, the bullet has its maximum velocity as the pressure that has built up during hte gunpowder explosion is no longer confined by the barrell and the bullet, it gets released to the atmosphere.  The bullet will then travel on a trajectory that depends on its weight, initial angle of travel relative to the horizon, and muzzle velocity.  There is a whole science dedicated to this field of study called ballistics.  In general terms, the bullet will start sinking as it leaves the gun but its forward motion is much greater than its downward motion so the first couple hundred yards of a high powered hunting rifle is hardly noticable and just a couple inches.  However, as the bullet travels further, the atmosphere slows it down in the horizontal direction but gravity speeds it up in the verticle direction and it starts to sink faster and faster until it hits a target or the ground.