What would be the speed of a bullet fired from a gun?
79521
It depends on the type of bullets and the type of gun that you use to fire the bullet.\n\n.22 rimfire cartridge 1200 - 1500 fps \n.22 centerfire cartridge 2400 - 3000 fps \n.22 Swift 4000 fps \n.38 Special 600 fps \n.221 Fireball 2650 fps \n\nThis gives you an idea of the velocity of bullets.
862116
KE = 0.5*m*(v)^2 = 0.5*0.023*(848)^2 = 8269.696 J\n\nThis must = work to accelerate the bullet\n\nW = F*d = 8269.696\n\nF= 8269.696/0.803 = 10298.5 N
332849
800 rounds per minute \ncheck out this site:\nhttp://www.hkdefense.us/pages/military-le/smg/mp5/mp5.html
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.
851042
Weight of the bullet..IE. grain  how far it'll travel @ what velocity
157080
This is just the regular projectile motion problem (see your text) just with an added initial velocity in the x direction (the velocity of the ship).\n\nLet's take as our positive direction the direction the gun shoots. (That is, the ship is moving in the negative direction.) Thus, the initial x component of velocity of the projectile is the x-component of the guns velocity less the velocity of the ship. (I'm going to use Vo for the initial velocity, since using "not" as a subscript confuses me.)  And I'll use A for alpha to save space.\n\nVxo = Vo cos(A)- v\nVyo = Vo sin(A)\n\nOur equations of motion are\n\nX = Xo + Vox t + 1/2 ax t^2\nY = Yo + Voy t + 1/2 ay t^2\n\nTo find the range, we want when the projectile splashes back down again, i.e. Y = 0.  If we set Xo = 0, then X will be the range, i.e. the X position when Y = 0 again.  And there is no acceleration in the x direction, the y acceleration is -g, the acceleration of gravity.  Plugging all that in, we get\n\nR = Vxo t \nR = (Vo cos(A) - V) t                     (1)\n\n0 = Vyo t - g t^2/2\nVyo t = g t^2/2\nVyo = g t/2\n2 Vyo/g = t\n2 ( Vo sin(A)) / g = t                     (2)\n\nPlugging our expression for t (Eqn. 2) into Eqn. 1 yields\n\nR = (Vo cos(A) - V)2 ( Vo sin(A)) / g \n\nwhich is the formula you gave.\n\nTo find the max range, we can use calculus.  There may be a more straightforward way if you don't know calculus, but this works for me.\n\nThe range formula is a function R of variable A.  To find the max, take the derivative and set it equal to 0.  Thus\n\ndR/dA = 2 Vo/g [sin(A)(-Vo sin(A)) + (Vo cos(A) - V)cos(A)]\n       =2 Vo/g [-Vo sin^2(A) + Vo cos^2(A) - V cos(A)]\n\nusing a trig identity\n\n       =2 Vo/g[Vo cos^2(A) - Vo + Vo cos^2(A) - V cos(A)]\n       =2 Vo/g[2Vo cos^2(A) - V cos(A) - Vo]\n\nWhat we have here is a quadratic equation with the variable cos(A).  When we set it equal to 0, in order to find the minimum, we can use the quadratic formula.\n\ncos(A) = (- (-V) + SQRT[V^2 - 4(2Vo)(-Vo)])/2 (2Vo)\n        = (V + SQRT[V^2 + 8Vo^2])/4Vo
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.
1157115
No, it falls back down at its terminal velocity(maybe), which is slower than the muzzle velocity because of drag.\n\nThe bullet leaves the gun with a given amount of kinetic energy: 1/2*m*v^2 where m is the mass of the bullet and v is its initial velocity.\n\nAir has non-zero viscosity, which is a resistance to shear forces.  When the bullet flies through the air, it drags air along with it.  This gives some of the bullet's energy surrounding air, reducing the bullet's velocity and kinitic energy.  Drag forces between fluids and solids occur in the direction opposite their relative motion.  (If the fluid is stationary and the bullet is moving, the drag is in the opposite direction of the bullet's motion.  If the fluid is moving and the object is stationary, the drag on the solid is in the direction of the fluid motion.)\n\nSimultaneously with drag, gravity is applying a force toward earth.  This too reduces the bullet's velocity and kinetic energy.  However, as the bullet flies higher into the air it gains an equal amount of potential energy from gravity.\n\nThe net result is that the bullet is always losing energy in flight because of drag.  This means that the amount of potential energy at the top is less than the initial kinetic energy, and it is thus impossible for the bullet to be going at the same speed it was shot when it hits the ground.\n\nThe fastest an object will fall under the influence of gravity is the terminal velocity.  At this velocity, the force of drag (which increases with the size of the object's profile, the viscosity and density of the fluid, and the relative velocity of the solid and the fluid) exactly balances the gravitational force.
59102
If the bullet is fired in the horizontel direction, both object will reach ground at the same time.. because they are reaching ground due to vertical gravitational acceleration only which is same for both.\n\nBut if the bullet is fired in the downward direction and at an angle downwards, it will reach grounf first because it is having a initial velocity as well whereas the initial velocity for the other object is zero.. The reverse holds true if the bullet is fired upwards.
903139
A bullet of course.  A gun uses a small amount of kinetic energy for the trigger hammer to strike the expolsive cap in the shell but thereafter the explosive force is sent into the bullet, with and equal opposite reaction energy dissipated by providing recoil which the gun uses (if automatic) to load the next shell.\n\nI assume the trick in the question is to ignore the equal and opposite reaction rule but not all of the force thus generated is in the form of kinetic energy which relates to bodies in motion.
170375
if you shot a gun from a car, does the bullet come out any slower??? no, the speed of your travel adds to the speed...
567093
9 mm Luger Parabellum\n\n* 7.5 g (115 gr) Full Metal Jacket: 1280 ft/s (390 m/s) \n* 8.0 g (124 gr) Full Metal Jacket: 1180 ft/s (360 m/s)
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?
1114215
17 HMR (Hornady Magnum Rimfire) ammo has a 17 gr. hollow point (filled with polycarbonate) boat tail bullet with a 0.172" diameter and delivers a muzzle velocity of about 2550 fps (depending on the barrel length). There is 5.4 gr of what looks like Lil' Gun powder. The muzzle energy is 245 ft-lb and with a 1-10" twist, the bullet is spinning at 183,500 rpm. A typical 22 Long Rifle cartridge has a bullet weight of 40 gr. or more than twice the mass of the 17 HMR.
887139
Momentum is the direct result of the following Newtonian Law of Inertia\nAny object in motion remains in motion and any object at rest remains at rest (unless a net force acts upon the object)\nMomentum (p) is the mass (m) of the object times its velocity (v).\np = mv\n\nForce (F) and Newtons Second Law can be derive from a change in momentum.  That is (t is time)\nF = delta p/delta t \nF = delta (m*v) / dt\nwe usually say m is constant so\nF = m * v/dt\nF =  ma    (a is acceleration or a change in v/time)\n\nNewtons third law is also based on momentum.  "For every action there is an equal, yet opposite, reaction."\n\nSo...  When a gun is fired what we see is the bullet propelled forward and the gun (and gun holder) propelled back.  Both received the same Force but because the gun and person holding it are much more massive than the bullet, the bullet will travel much faster. \ne.g F1 is the force on the bullet which is the same as the force on the person F2\nF1 = F2\ndelta p1/dt = delta p2/dt\nm1v1/dt = m2v2/dt\nm1v1 =m2v2\nImagine the bullet is 1 gram and a 180 lbs (~8182 grams) person shot the bullet...  The avg v on the kick back will be about 1/8182 of the v of the bullet...  Lucky for the shooter.
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
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?