what happens to the bullet when someone shoots into the air?
407036
The bullet will eventually fall back to Earth, however, it will not be going nearly as fast as it was once shot out of the gun.\n\nThe muzzle velocity of a bullet out of a gun is much higher than the terminal velocity of that same bullet falling down through our atmosphere.  As soon as that bullet leaves the barrel, it begins to slow down due to air resistance.  As the bullet climbs higher and higher, gravity too is working to slow the bullet down.  The bullet will reach its highest point and begin to fall back down.  Gravity works to accelerate the bullet downward, but air resistance works to slow it down. The bullet will reach some maximum downward velocity when the force of friction (air resistance) equals the bullet's weight (force of gravity) and will not accelerate any more.  The bullet will hit the ground traveling much slower than it did when it was fired from the gun.\nThat said, I still wouldnt want to be the person it falls on before I do any terminal velocity calculations for the particular bullet.
944006
a) Initial velocity, u , of the bullet as it hits target=1200m/s\n    final velocity, v, as it emerges =850m/s\n    distance traversed,s = thickness of target\n                                   =30cm\n                                   =0.30m\n\nv^2-u^2=2as\n\n850^2-1200^2=2a*0.30\n722500-1440000=0.6a\n-717500=0.6a\na=-717500/.6\n  =-1,195,833m/s^2\n\nv=u+at\n\n850=1200+(-1,195,833)t\n1,195,833t=1200-850\n               t=350/1,195,833    \n                =2.93*10^-4  s\n\n              \nb) F=ma\n mass of bullet, m=25grams or 0.025kg\n\nF=0.025*1,195,833\n  =29,896N\n\nc)Initial momentum of bullet=25*1200\n                                         =30,000\nInitial momentum of target=350*0\n                                          =0\n\nFinal momentum of bullet=25*850\n                                         =21,250\nFinal momentum of target=350v\n\nBased on conservation of momentum:\n\nInitial momentum of the system=Final momentum of the system\n\nSubstitute known values:\n\n30,000+0=21,250+350v\n350v=30,000-21,250\nv=8750/350\n  =25m/s
808390
Bound's hubby here:\n\nTechnically, a bullet begins to drop the moment it leaves a rifle's (or pistol's) barrel. It appears to rise because of the angle of elevation the firearm is at in order to hit a distant target. \n\nNow, there can be a negligible amount of rise with a firearm whose barrel has a right-hand twist firing into a wind coming directly from the right (and vice versa). This is due to the rotation of the bullet caused by the barrel's twist and the bullet's velocity and its interaction with the wind. However, this amount of "lift" is so insignificant, it is practically non-existent.
963941
Bound's hubby here:\n\nFirst off, how certain are you that the bullets are penetrating the bowling ball and not ricocheting off the round surface of the bowling ball? Second, you should be aware that the core of the bowling ball is almost solid cork, which will absorb more shock than you can dish out. Your bowling ball is really a poor choice in target from a safety stand point.\n\nGood luck.
1024308
Friction aside, think of it like this:\nwhen the bullet is fired it has a certain amount of speed.  every second that speed is being reduced by gravity by 9.8 m/s.  So when it reaches as far up as it will go it stops because gravity has subtracted 9.8 from from the original speed every second until the speed is 0.  Like taking out money from an account.  So now you can spend all the money you have-it's got potential.  Then it starts moving toward earth.  gravity is still working at the same rate. so it picks up speed at the same rate that it lost it last time. And it's the same distance, so it'll take the same amount of time.  So now you're depositing the money into a different account.   You only have the money you took out of the first account, so that's how much the second account will have.\nI hope that explains it for you.
200339
Conservation of momentum.  Newton tells us that for every action, there is an equal and opposite reaction, such that in a collision, the net momentum remains the same.  When you realese gas backwards, the momentum imparted on the gas away from the space craft needs to be ofset by imparting momentum upon the astronaut in the direction of the space craft.\n\nWhen you fire a gun, the momentum of the bullet forward is equal to the momentum of he gun backward.  Since the gun is much bigger than the bullet, it doen't move as fast, and the force is spread over a larger asrea (your hands and shoulder).  But the kick backwards is the same sort of principle: the equal and opposite reaction.
710427
You've got some good answers here, espeically the guy with the links, but my two cents.\nSolid, non expanding bullets, such as most military and police ammunition tend to go "through and through" the target when they hit it.  Since the bullet goes out the other side basically undamaged, it transmits relatively little of its kinetic energy to the target.\nSoft-point bullets do a little better, flattening on impact to present a broader wound track and transmitting more energy to the target.\nHollow-point bullets take this a step farther, the thin walls of the bullet around the hollow (in theory) peel back, expanding the bullet faster than a soft-point bullet does - with a corresponding increase in energy transferred to the target.  In addition, many hollow point bullets fragment, resulting in multiple wound tracks and more rapid blood loss.\nRemember, when you're hunting the critter you've shot, unless it is very small, is not likely to die instantly of the shock of being shot.  Unless you make a head shot (which hunters generally do not do due to high miss probability), the animal is going to run until it bleeds out.  The larger the wound, and the more energy transferred from the bullet in flight to the animal, the faster and more humanely this happens.
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
965199
Firing the weapon vertically...\n\nV^2 = Vo^2 + 2gh\n0 = 600^2 + 2(-9.81)h\nh = 18.348 km ; where h is the maximum height reached by the bullet\n\nThe work done by firing this bullet to reach the max height is...\n\nW = mgh\nW = (5 gf) (9.81 m/s^2) (18.348 km)\n\nBTW, 5 gf? What the fcuk is that? I'll just assume this as 5 grams.\n\nW = 900 N-m or Joules\n\nPower is the rate of doing work. Power therefore is equal to...\n\nP = W x bullets fired per min.\nP = 900 x 250 bullets / 60 secs\nPower = 3,750 Joules per sec or watts.
221534
If this is true, then you may be aware of the "fact" that if someone shoots an arrow at you, it will never hit you, because after a certain amount of time after being shot, the arrow is halfway to you; then more time goes by and the arrow is half of that distance, then more time and half of that, etc. so there is always half of some distance left and some time left so it never hits you.
939061
The base of the bullet contains a small bit of phosphorous which is ignited by the burning gases from the ignition of the cartridge. This burns for as long as it takes the bullet to travel @400 meters. The tracer effect can be used under low light conditions to show you where your bullets are impacting (and the enemy to see where your bullets are coming from), allow you to 'designate' targets visually, and (to a limited extent) act as an incendiary device on inflammable materiel.
420139
Not unless the wind carries it because whilst the earth is spinning around rapidly, so is everything on it. If this were not the case, and a person jumped straight up into the air they would shoot of in a direction and this doesnt happen
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.
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.
901927
Recoil, in common everyday language, is considered the backward kick or force produced by a gun when it is fired. In more precise scientific terms, this force is equal to the derivative of the backward momentum resulting when a gun is fired.\n\nThe backward momentum is equal to the mass of the gun times its reverse velocity. This backward momentum is equal to the sums of the two forward momentums by the law of conservation of momentum and is due to and equal to the projectile's mass times its velocity added to the mass of the escaping gases, burnt propellant, unburnt propellant, and other gases, times their respective velocities. \n\nFor your case, we don't know enough information about the gases and other by-products of the firing. However, if we assume they are negligable, we can focus on the bullets.\n\nConservation of momentum:\nMh * vh1 = Mh * vh2 + Mb * vb\n\nwhere;\nMh is the mass of the helicopter\nvh1 is the initial velocity of the helicopter\nvh2 is the velocity of the helicopter after the firing\nMb is the mass of the bullets\nvb is the velocity of the bullets\n\nDepending on the accuracy of the answer you want, you may want to consider that the mass of the helicopter will change as the bullets leave the gun.\nMh * vh1 = (Mh - Mb) * vh2 + Mb * vb\nhowever, you don't have enough information to solve it in this case. You would need to know the velocity of the helicopter after the firing (vh2). Don't worry about it, I am probably just confusing you.\n\nBack to the question:\nMh * vh1 = Mh * vh2 + Mb * vb\n\nMh * vh1 - Mh * vh2 = Mb * vb\n\nMh * (vh1 -  vh2) = Mb * vb\n\n(vh1 -  vh2) = Mb * vb / Mh
996188
If the risk of removing the bullet from the patient out weigh the benefit, they will leave it in. \n\nie: next to the spine, spinal cord...
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?