The faster a bullet travels, the less powerful it is? I don't get it.?
487788
you're talking .223 here aren't you.  the heavier bullet will retain more energy than a lighter bullet.  imagine getting hit with a ping pong ball, a marble and a ball bearing, each traveling about 100 mph.  the ball bearing will hurt more because it has more energy in its weight.\n\nhowever, the benefits of a lighter bullet is that it will go faster and will drop less at shorter to medium ranges.  the ligher bullets are good for varmint shooting.  the heavier bullets are good for larger animals.
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.
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
810426
From your point of view on the plane (as well as the point of view of any other aircraft traveling the same speed and the same direction), the velocity of the projectile will appear to be the same whether fired forward or backward. However, the velocity will drop off faster when fired forward from the plane than it would on the ground due to the increased wind resistance produced by the head wind.\n\nFrom the point of view of someone on the ground, the speed of the bullet would be added to the speed of the plane when fired forward and subtracted from the plane when fired backward.\n\nUltimately, the velocity of any object is relative to the reference frame of the observer.
372578
As fast as you can shoot it dumb***
120276
The first thing you have to understand is that the speed of light in a vaccum is constant, no matter how fast you're moving, nor how fast the source of the light is moving. This is bizarre. If someone standing on a conver tosses a ball to you at 5 mph while you're standing on the corner, you'll see the speed of the ball as being 5 mph relative to you. But if you're driving at 15 mph and they throw it at you at 5 mph, the speed of the ball relative to you will be 20 mph. This is how most things work. But with light, no matter the speed of the source or yourself, you'll always measure the speed of light as being 299,792,458 metres per second.\n\nEinstein accounted for this by postulating his special theory of relativity, which says basically that the reason for this is because if you move really fast, three things happen: 1) time slows down, 2) mass increases, and 3) length in directions parallel to the direction of movement contract. If you take those three hypotheses, you'll also note that the concept of things being "simulatanous" is a relative term, meaning two observers moving at the same velocity (or moving at no speed relative to one another) will agree on which events are simultaneous and which are not. However, observers at different velocities will have different concepts of what is and isn't simulataneous--and yet THEY'RE BOTH CORRECT. It's pretty mind-blowing, but if you take all of these notions together, you get a consistent view of reality.\n\nThat's Einstien's SPecial Theory of Relativity. About a decade later he formed his far more complicated general theory of relativity.\n\nThe general theory says that Newton's view that gravity operates by exerting a force on a falling object is incorrect. Rather, by general relativty, a free-falling object that is in a vacuum has no forces acting on it at all. The movement of falling objects isn't due to force, but rather the curvature of space-time--massive objects actually curve the surrounding universe. In Newtonian physics, we regard there being three spacial dimensions (up/down, left/right, and backwards/forwards) and a single dimension of time. In general relativity, there are four dimensions of space-time. In Newtonian physics, a foce makes the earth orbit around the sun. In general relativity, the earth and the sun curve the surrounding space-time continuum, and both bodies move in a straight line through curved space, resulting in the perceived sense that one body is circling about the other. In Newtonian physics, light should not bend to gravity, because gravity affects massive bodies, and light has no mass. But in fact, light does bend around stars and such.\n\nGeneral relativity has been likened to a stretched sheet of elastic material, like a trampoline. Heavier objects sink more, and make the sheet curve about them, and light obejcts rolling across are deflected by this curving.\n\nGood luck understanding all that. Hope your head doesn't explode.
325948
It doesn't translate.  Some people are very quick...get to full speed in a hurry.  Others take more time, but reach faster speeds.  It is much like the Quarter Horse vs Thorobred.  Have I confused you enough?  I could go on....
294744
How big of a bullet?\n\nIf we're talking standard handgun round, It would definitely hurt, but there's not 100% certainty that it would kill them. The terminal velocity of a bullet free-falling is not nearly as fast as a bullet shot from a gun, due to the lack of spiral flight pattern and the wind resistance involved in such a fall.\n\nIt would basically be about the same as dropping a penny off the empire state building, which has been shown by Mythbusters that it won't kill you, but will certainly sting.\n\nNow, if we're talking larger ammo, i think we're definitely getting into the range of being lethal.
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.
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.
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.
960079
You've lost the point of the equation.  Squaring the speed of light will give you the energy released, not an enhanced speed of light.
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.
296770
i suppose it depends on the car.  there are a few cars that are jet powered that can travel faster than the speed of sound.
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
744405
a) (speed is actually closer to 3.97...not a big deal)\n\nb) Conservation of energy:\nThe sum of potential and kinetic energies when the bullet first hits the block and when it is at the top of the swing is the same:\nU(bottom) + K(bottom) = U(top) + K(top)\nNow, the speed at the top is zero, so the ewuation becomes:\nU(bottom) + K(bottom) = U(top)\nSolve for the change in potential energy:\n[U(top) - U(bottom)] = K(bottom)\nThe change in potential energy due to gravity is: (m1+m2)gh\nThe kinetic energy at the bottom is: (1/2)(m1+m2)v²\nSo the equation becomes:\n(m1+m2)gh = (1/2)(m1+m2)v²\nThe (m1+m2) terms cancel:\ngh = (1/2)v²\nSolve for h:\nh = v² / (2g) = 0.8 m
786520
The muzzle velocity of the paint ball gun is higher. \nhttp://en.wikipedia.org/wiki/Muzzle_velocity