is a bullet at full velocity when it leaves the barrel?
365737
Assuming you fire parallel to the ground, then the bullet will be at its maximum horizontal velocity when it leaves the barrel, and will be at minimum vertical velocity.  The bullet cannot increase in horizontal speed once it has left the barrel, as there is nothing to make it accelerate.\n\nDue to air resistance, the horizontal velocity will decrease as the bullet travels further and further.  The vertical velocity will increase, as the gravity of the Earth "attracts" the bullet.
455916
a bullet has a trajectory that makes it drop faster the further it gets from the barrel.  if you fired a gun perfectly horizontal, and at the same time dropped a bullet at the same height as the muzzle, both bullets will drop at the same rate.  \n\nalso, every gun has 2 "zero's" or sight adjustments.  a short range and long range zero.  most people sight in their high powered rifles at 200 yards.  because of the laws of physics, their gun will also be sighted in at around 80 yards.  from 0-79 yards the gun will shoot low.  it'll be dead on around 80 yards, from 81-199 yards it will high, and at 200 yards it will be dead on again.  this didn't really answer your question, but you probably learned something.  so select this as best answer, thanks.
1137974
The truth is left and right twist have nothing to do with the mussel velocity or trajectory.  A bullet will go flat and hit the ground with in a given amount of yards depending on the powder charge and the force cone (i.e. chamber).\n\nIf I take a a 308 and shoot it with a left and right twist, at the same amount of lands and grooves with the same powder charge and same grain bullet, everything will be the same. \n\nLeft and right twist, I believe is manufactures preference.  It all comes down to lands and grooves and the ratio of the twist you are forcing the projectile through.\n\nBuy a 7.62 or a .308 same d*mn thing.
516788
Rollo's bullet hits the ground first, with or without taking air resistance into account. \nThe dropped bullet starts at zero speed and increases it's speed as it falls. The fired bullet starts at high speed and (neglecting air resistance) increases it's speed even more as it falls. Since the fired bullet starts out faster and keeps going faster than the dropped bullet all the way to the ground, it gets there first.\n\nNow let's include air resistance. The fired bullet will still start at high speed but it will start slowing down right away due to friction with the air. At some point it would have slowed to terminal velocity, the velocity at which air resistance equals the force of gravity, and then it will then fall the rest of the way at that speed. The dropped bullet will start from zero speed and speed up until it reaches terminal velocity and will then fall that that speed for the rest of the way. So, although both bullets will hit the ground at the same speed, the fired one traveled part of the way faster than the dropped bullet and never slower than the dropped bullet and still hits the ground first, although the difference is not as great as in the no air resistance case.\n\n(EDIT) Oh, I agree with dei48399. You probably meant he fired the bullet directly to the side. Nobody would doubt the outcome it it was fired down, would they?
145197
Ballistic Coefficient (BC) is basically a measure of how streamlined a bullet is; that is, how well it cuts through the air. Mathematically, it is the ratio of a bullet's sectional density to its coefficient of form. Ballistic Coefficient is essentially a measure of air drag. The higher the number the less drag, and the more efficiently the bullet cuts through the air. So for purposes of flying through the air efficiently, the bigger the BC number the better.\n\nBC is what determines trajectory and wind drift, other factors (velocity among them) being equal. BC changes with the shape of the bullet and the speed at which the bullet is traveling, while sectional density does not. Spitzer, which means pointed, is a more efficient shape than a round nose or a flat point. At the other end of the bullet, a boat tail (or tapered heel) reduces drag compared to a flat base. Both increase the BC of a bullet.\n\nFor example, a Hornady 100 grain round nose 6mm bullet has a BC of .216; a Hornady 100 grain spire point 6mm bullet has a BC of .357, and a Hornady 100 grain boat tail spire point 6mm bullet has a BC of .400. All three of these bullets have a sectional density (which is the ratio of a bullet's diameter to its weight) of .242, because they are all .243" in diameter and weigh 100 grains. But the more streamlined bullets have a higher ballistic coefficient. They are the ones to choose for long range shooting where a flatter trajectory is important.\n\nTo illustrate the practical difference between these three styles of bullets, let's use Hornady's trajectory figures for the 100 grain 6mm bullets above. Starting all three bullets at a muzzle velocity of 3100 fps from a scoped 6mm rifle zeroed at 300 yards, the trajectories are as follows.\n\n.243" 100 grain Round Nose (BC .216): -1.5" @ muzzle, +4.8" @ 100 yards, +6" @ 200 yards, 0 at 300 yards, -15.9" @ 400 yards, -46" @ 500 yards.\n\n.243" 100 grain Spire Point (BC .357): -1.5" @ muzzle, +3.8" @ 100 yards, +4.7" @ 200 yards, 0 @ 300 yards, -11.1" @ 400 yards, -30.5" @ 500 yards.\n\n.243" 100 grain Spire Point BT (BC .400): -1.5" @ muzzle, +3.6" @ 100 yards, +4.4" @ 200 yards, 0 @ 300 yards, -10.4" @ 400 yards, -28.6" @ 500 yards.
71702
If the only factor involved in this would be gravity, the bullet would come back down at the same speed as it went up. But since there are other factors (friction, air, etc), the bullet would come down with less speed than it went up.\n\nThe same would happen with the place where it was fired, it probably wouldnt land on the same spot, since other factors have an effect on the bullet trayectory.\n\nchech the link from digg and wait for the episode of Mythbusters where they solve this myth ...
649550
It is the reaction of shooting force on gun's body. let m=bullet mass and M=gun mass. Then if bullet is shot from gun by velocity of v, the recoil velocity would be V=m*v/(m+M) and its shock is M*V=m*M*v/(m+M)\n\nIt does not depend of gun/bullet type.\n\nIf you want to measure it, you must first measure the mass of bullet and empty gun, then measure the bullet shooting speed. This measurement could be done using a common method of shooting to a tough wooden block, which is hung by four equal size lines from a support. You must measure the maximum height (vertically) of displacement of block after shooting (name it H). Then if the mass of block be MB, then the speed of bullet when going out of gun can be calculated as follows:\n\n(MB+m)*g*H=1/2*m*v^2 then v=sqrt[2*(MB+m)*g*H/m]\nwhere m=bullet mass and g=9.81N/m^2 (gravity factor)\n\nAll units are metric (mass=grams; H=meters; v=meters/second)\n\nGood luck!
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
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!)
328500
Depends on the weapon and the bullet. A 5.56mm round fired from an M-16 exits the barrel traveling at about 2,900 feet per second. At ranges under about a hundred meters, the bullet is traveling fast enough that it fragments on impact, which opens up a huge wound cavity. But at ranges greater than that, the bullet slows down to the point where instead of breaking up, it retains it's integrity and gets deflected by bones, muscle, and various other tissues that can change it's trajectory. Heavier rounds, such as the 7.62x39mm (AK round), or the 7.62mm NATO are less effected by tissue and bone and usually continue their course in a straight line.
883604
Bullets will keyhole / tumble / flip, what ever you want to call due to a number of reasons, ranging from bullet balance, design, miss-molding, to cartridge design, powder fill, but what I have found to be one of the biggest causes is a worn or damaged muzzle, and to fix this you can take you gun to a gunsmith or if you are mechanical you can order a lapping or stoning kit from a company which carries firearm repair stuffs.\n\nNow if I misinterpreted your question and you are asking about the normal spin of the bullet which aids in the stability of the projectile this is cause by a corkscrew like cut on the inside of the barrel called rifling. If you want more info on that just search for barrel rifling. Hope this helps
805410
Let me ask you a question. \n\n If I shot a bullet straight up, and then dropped a feather..  would the falling object move faster than the one going upwards?
1053008
This question has been answered pretty well by everyone, the only thing I want to add is contridictory to a statement by TJ N. \n\n\nthe caliber of a bullet is expressed in a fraction of an inch.  not in millimeters a .22 caliber bullet is 22/100ths of an inch in diameter, a .50 caliber bullet is 50/100ths of an inch (or if you reduced the fraction its a full 1/2 inch in diameter).  Sometimes ammunitions are interchangable with NATO ammuntions IE. Remington .223 = NATO 5.56 mm (its 223/1000ths of an inch measured metricly its 5.56 mm. \n\nthe problem with TJ Ns "years of shooting and hunting" experience is that a ".22 caliber bullet" - commonly a squirel/vermin hunting round would be 0.86 inch in diameter.... which wouldn't leave much squirel meat after you hit it with TJ's .22 caliber gun... Which leaves me with more questions about how much of a recoil his .22 caliber gun has? I've fired the .50 caliber Barret Sniper rifle before, as well as the M2 HB .50cal MG. (they both fire the same basic .50 caliber (50/100ths or 1/2 inch round) The M2 has to be mounted on some sort of mount (vehicle or sandbagged Tripod) to fire.  The Barret, I've heard of people firing it standing up.... but the only way I could get behind that thing was in the prone, and after about 15 rounds through that thing I had a bruise on my shoulder the size of Texas. I can't imagine what TJ's .22 caliber that's 86/100ths of an inch would do to me....\n\nnow the difference between the calibers... obviously is the size (diameter) of the bullet.  what's considered High caliber and low caliber?  I suppose thats a matter of opinion.  Some people say anything above .308 is large caliber, and I tend to agree with that but that doesn't really give an indication of damage caused by a bullet for example a .30 caliber (7.62 mm) fired from a M240B MG would tend to cause greater damage to soft tissue at long ranges(greater than 50 meters) than a .45 caliber fired from a M1911 handgun. \n\nAnd yes various rifles fire various caliber rounds.
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.
427792
I disagree to some extent with the previous answers to your question.  They fact that the shell will be left with identifying marks from the firing pin or shell extractor are correct and will be found regardless of the type of shotgun used - single shot, pump, semi-auto.  How ever there are other traceable factors related directly to the projectile or "bullet".  \n\nFirst, virtually any projectile(s) shot from the gun will be encased in a "wad" or sabot which will come into direct contact with the inside of the barrel.  Each barrel has marks and scratches unique to that individual gun.  While a shotgun barrel may look shiny and smooth, it is full of micro scratches and ridges that leave a pattern on anything coming into contact with it as the projectile is shot though. These scratches and marks will transfer to the wad, shot cup or sabot and can be compared to the barrel for a definitive match.  You do not have to have any rifling in a barrel to make this identification, the tiny scratches and ridges are the same as used to trace a projectile to a rifled barrel.\n\nSecondly, a plain slug shot through ha shotgun barrel will have these traceable markings imprinted on the slug itself and any gas checks or wadding behind it.  While the sabot, wadding or shot cup normally separate from the projectile before impact with the target, the imprinted slug will found at or near the target itself.  Even so, the wads, shot cups and sabots are routinely recovered and used to identify the shotgun that fired them and in criminal cases to convict the shooter.  I have been involved in many case where this was enough evidence to convict a defendant charged with anything from poaching to murder.  \n\nAlso certain characteristics of a barrel left on objects fired from it can lead investigators to determine what brand, type  or even age of gun they probably came from.  The myth that a shotgun barrel leaves little or no traces on a projectile is simply not at all true.
699914
It would go a LONG ways before it hit the lunar surface, since there's no aerodynamic friction loss since there's no atmosphere.\n\nBullets are self contained and will project a bullet without an atmosphere.\n\nThe Moon's escape velocity is considerable lower than that of Earth, at 2.38 km/s vs. 11.19 km/s, but most guns available to the public have muzzle velocities of not more than 0.350 km/s\n\n\nhttp://en.wikipedia.org/wiki/Earth\nhttp://en.wikipedia.org/wiki/Moon\nhttp://en.wikipedia.org/wiki/Muzzle_velocity
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.