If shooting a bullet 90 degrees upward sky where should it fall (while no winds around)? When it falls?
478202
If is really no wind, it should fall right back into your gun.\n\nBut an experimental proof I think can hardly be done because of friction effects between air and bullet: so you would need an evacuated tube of some 100 m height, higher than any building. And the accuracy of this tube also woud be hard to be realized.\n\nNevertheless, such an accuracy problem I think was encountered when was built the SLAC (STANFORD Linear Accelerator) in the 1950ies; but this installation was not upright.
1046911
it will fall straight down , but appear to fall at an angle if observed from the plane
375814
depends on gun speed, wind speed, and the angle it has been shot at. my friend his his friend fron 120 yards away with a stock gun like the ones you find at walmart
1137906
depends on where you were shot
923444
Short Answer: \n\nvert comp = .161 meters\nhori comp = .936 meters                        \n\nLong Answer:\n\nMb=mass of bullet\nMp= mass of pendulum\nv=velocity of bullet\ng=9.81 m/s/s\nh=vertical displacement\n\n\nBy conversation of momentum:\n\nmomentum of pendulum with bullet=momentum of flying bullet\n\nsqrt(2*g*h)(Mb+Mp)=Mb*v= (28/1000)*230=6.44 \n\nsolving that you can find h=0.161 meters\n\n\nBy the distance formula:\n\nsqrt(2.8^2+0^2)=sqrt( (2.8-0.161)^2 + (horizontal position)^2)=2.8\n\nsolving that you can find horizontal position = 0.936 m
826715
bullet drops 0.02 m due to gravity\n\nfree fall distance formula:\n\nd=v1xt + 0.5xgxt^2, v1=0 ( initial vert velocity is 0) and d=0.02 m\n                                   g=9.81\n\nt= (2x0.02/g)^0.5 = 0.0639 s\n\nv=25m/.0639= 391.5 m/s
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.
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.
925506
You need to consider conservation of momentum and conservation of energy.\n\nConservation of Momentum says that the momentum before tha collision is equal to the momentum after the collision.\n\nm*v1 = (m + M)*v2\n\nwhere m is the mass of the bullet, M is the mass of the block, v1 is the speed of the bullet, and v2 is the speed of the bullet/block system. Therefore,\n\nv2 = m*v1/(m + M)\n\nTo find out how high the bullet/block system rises, consider conservation of energy. The kinetic energy of the bullet/block is converted to potential energy as the system rises so that\n\n0.5*(m + M)*v2^2 = (m + M)*g*h\n\nwhere g is the gravitation acceleration and h is the height to which the bullet/block rises. Solving for h\n\nh = v2^2/(2*g).
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
248915
The limit as sky goes to zero from the positive direction would be ground.\nLim    = ground\nsky-->0+\n\nThe limit as sky goes to zero from the negative direction is space.\nLim    = space\nsky-->0-\n\nSince the limit from the + site does not equal the limit from the - side, the limit does not exist.
692673
Outward from the source at first.  As it becomes larger, it moves in the same direction as the wind is blowing.  Fires also move more readily up a slope than they do down a slope.
19099
The knob on the side is your elevation, "or Up & Down". The Knob on top is your windage. "Left & right.\nTake 3 very steady shots at a target, not too far away. Look at your target, mark the holes with a pen or cover with tape or something so you do not loose them. Then raise of lower you elevation. Then 3 more shots... then adjust, over & over again. You want to hope that the 3 shats are always together, as you make each adjustment. Once you get the elevation, Next work on you windage. Turning the knob left, moves the shots to the right, & vice versa. You really do not want to have your target too far away in the beginning, because you need to get a tight 3 shot pattern to be able to make adjustments. once you are on target at say 25 or 30 yards, then move the target farther away. In the beginning, you just want that tight 3 shot pattern to make accurate adjustments. Don't forget to hold your breath & squeeze slowly. You will get a stray shot from time to time, but this can be expected. Takes about an hour to do it right.
1111942
Not necessarily. If your theoretical gun shoots a bullet that reaches escape velocity [1] (the speed something needs to travel to beat Earth's gravity), it can overcome the Earth's gravity [2] and fly off into space.\n\nHowever, I know of no such gun. Only rocket engines [3] are able to do this, because they sustain and increase their speed by burning fuel. A bullet has nothing to keep its speed up once it leaves the gun.\n\nSo the answer is, "Yes, the bullet comes back down." The bullet is always beholden to gravity, even before it's fired. You can view gravity as an invisible string always pointing to the center of the earth, always being pulled. The instant the bullet leaves the barrel of the gun, it will be constantly slowing down due to gravity (and air resistance).\n\nAnother interesting idea is this: If you were to take a gun and fire it horizontally, and at the same instant drop a rock from the same height as the gun, which would hit the ground first? The bullet or the rock? \n\nThe answer is that they both would hit the ground at the same time. The bullet would just be several hundred feet away when it hits. Both are affected by gravity's pull, and approach the ground at the same speed.
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
454546
never shooot a gun in the air. a bullet never has a name on it!! i wouldnt even shoot at the ground, it could bounce off a rock or something. stick to targets, much safer!
584655
45 degrees is the best angle in theory.\n\nIn practice you may have a adjust a little to compensate for air resistance and wind factor.