If you where to take any handgun or rifle and fire the projectile from the firearm at what point would the projectile reach it's maximum speed/velocity.
123457
muzzle, at which point there is no more force pushing the projectile and the point which air starts to slow the projectile down.\n\nblack powder burns slowly the length of the barrel, pushing the projectile with increasing speed until it leaves the barrel.\n\nSmokeless powder burns rapidly.
557958
1) No velocity can be perpendicular to acceleration at the point where the movement would be circular.\n\n2) When you talk about a projectile his vertical velocity is 0 at the highest point of his trajectory but not the acceleration, this one if you do not consider air resistance is constant at all times.
460430
first thing u should do is do draw the path of the projectile,  using only imagination, the ball will definetly fall down as it moves away from the table edge until it finally hits the floor.\n\nprojectile path is always parabolic, and the direction of the velocity vector is tangent to the path of the projectile at any point.\n\nat the initial state, the velocity vector was pointing left or right according to your assumption, but as the parabolic path falls down, the velocity vector points down with it, when the ball hits the floor the velocity vector will be perpendicular to the horizontal in the downward direction.\n\nso at 45 degrees downward, the velocity vector has two components, on in the horizontal V*cos(theta) and another downward V*sin(theta), to get how high the ball at this instant apply the same laws but use V*sin(theta) instead of V ,  and to find how far from the table use V*cos(theta) instead of V.\n\nhope this helped.
407978
With or without gravity, there are two types of friction, static friction from the current and voltage and kinetic friction of the projectile as it travels down the rails. The kinetic friction mainly comes from the projectile "scraping" across the rails. This results in rail erosion when firing, ejection would produce hot plasma and damaged rails. \n\nThus, gravity has little or no effect, unless you position your rails such that the projectile is resting on one of the rails, increasing the kinetic friction. This problem is resolved by experimenters, who put the projectile side by side with the rails.\n\n        []o[] <---- Cross-section of rails with a projectile in between\n\nOf course, the spacing in between the rails and the projectile yet again must be very close, so even if there was a region of zero gravity, it would be hard for it to shift.\n\nReduction of friction, at most, is a good rail surface, often polished for a long time and a well-shaped projectile. To help with the acceleration and speed of the projectile, and slightly reducing rail erosion by eliminating some skin friction, gas injections are also used to first push the projectile then quickly fired together with the voltage.
625910
a) It will decrease in speed as soon as the accelerative force of the explosive gas ceases to be greater than the drag of the gun barrel or the air or other media it passes through. This could mean that it begins to slow down in the barrel, or soon after leaving it. Certainly, after a few feet from the gun barrel one would expect the gases from the gun to be dissipated and the bullet to begin to be dragged by the air. However, the reduction in speed of a high velocity bullet could be negligible for seconds after firing.\n\nb) As soon as the bullet leaves the barrel it will begin to drop towards the earth. The formula for this is: \ndistance =0.5*acceleration*(time-squared)\n\nAcceleration at sea level is 9.81 metres per second per second. Thus it will take almost exactly 2 seconds for the bullet to drop the 20 metres from your firing platform. (square root of [20/(0.5*9.81)]  )\n\nc) See b above. The bullet is acted upon by gravity at all times. It is only prevented from being accelerated downwards by a force equal or greater to its weight pushing it up. In your example the gun barrel supports it until the bullet leaves the barrel, then begins to fall. Even if you pointed the gun straight up, the bullet will be slowed by gravity as soon as it leaves the barrel and gases behind, gradually slowing to a stop and falling back to land on your head.\n\nIf the bullet had enough horizontal velocity to make it to the horizon within two seconds, it would have acheived the required "escape velocity" to leave or orbit the earth, always falling but never landing. As the horizon at sea level is eight miles away this would need a speed of 28,800 miles per hour or 46,340 kilometers per hour. \n\nThe speed of sound is c.1,220 km/hour, so this would be 38 times the speed of sound!\nThe muzzle velocity of most military calibres is c.700metres/second or 2, 520 km/h or twice the speed of sound.\n\nThus in the two seconds it takes the bullet to fall to earth it would have carried some 1,400 metres.\n\nd) Very much so!
549492
Projectile Motion-  Examine the initial velocity and angle necessary to hit targets at different points.
578143
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.
386672
Here's the best site I have found for calculating the velocity of the spud:\nhttp://ffden-2.phys.uaf.edu/211.fall2000.web.projects/J.%20Jones%20and%20T.%20Kenny/Projectile.htm\n\nThis is for a combustible firing chamber.  The pneumatic guns are more predictable and reliable since the pressure can be set prior to firing.  Where the combustible require calculating the air to fuel mixture based on the size of the firing chamber.  This is explained on another part of the site referenced above.\n\nThe best choice for this type of gun seems to the the solenoid valves that are common on sprinkler systems.\n\nThey also recommend that the barrel resistance be measured using a simple fish scale to find the force of resistance of the spud against the barrel.\n\nI also looked at a rifled barrel that provides better accuracy.\n\nMost spud gunners use parametric measurements to find the range of their cannon.  This is done using inclimeters for altitude and pressure gages to measure the max pressure in the firing chamber, while increasing the amount of fuel or air pressure and taking a series of measurements to find the characteristics of the cannon.\n\nenjoy\n\nj
525888
Depends on the bullet and its grain.The average velocity of a 12 pdr projectile was around 900 fps with a range of \n1800 yards. ...  Field Howitzer shell
631653
There are some revolvers in mass production that use rifle rounds in them.  While they are firing a 45-70 rifle cartridge they are still hand guns and give repeat shot capability.  Link below will give you more info on this gun.
60293
Well I wont solve it for you \nbut I will help you. \n\nThese problems always involve the equation \ns=s0+v*t+(at^2)/2\n\nwhere s is displacement\nv is speed\nand a is acceleration\n\nNow you will always have acceleration of gravity g\n\nbut when the freak comes out of the canon, they will have a certain velocity. \n\nThey will be accelerated as the spring releases its energy, and then leave the barrel with a constant velocity, at that angle. \n\nThey will have a horizontal velocity and a vertical velocity. \n\nso you will need two equations\n\nx= Vx *t + .5*Ax*t^2\n\nand y=Vy*t+.5*Ay*t^2\n\nOne for vertical travel, and one for horizontal travel\n\nThe horizontal travel will will assume Ax=0 (not really true, but close enough)\n\nThe vertical travel, assume Ay=-g, acceleration of gravity. \n\nTo find the speed out of the barrel at the start, \n\nremember Hooke's law\n\nF=kx. \n\nSo you know k and you know x, so you know F\n\nYou also know Newton's 2nd law, F=ma.\n\nSo you know a =F/m\n\nNow you have to find the velocity V. \n\nAs the spring releases, it accelerates the freak with acceleration a. The only question is, over what distance is this acceleration acting? \n\nWell one idea is that it acts over the compressed distance. Another idea is that it acts over 2 times the compressed distance, since it will oscillate in approximately simple harmonic motion, and will continue to push that entire distance. I would guess the latter.\n\nSo put this in \n\ns=.5*at^2, \nand compute t\n\nFrom t, you can figure out the speed at launch: \n\nV=a*t\n\nSo you have the initial velocity V, which you can resolve into vertical and horizontal components, and then figure out how long it takes the freak to return back to the ground, and in that time, how far will the freak travel horizontally. \n\ngood luck\n\n- - - - - - - - - -\n\nkirchwey is correct. The same force is not applied through the entire distance that the spring oscillates. The force F=kx is the maximum force, only at the beginning of the freak's acceleration. I was thinking about that so I returned to this problem (it has been a long time since i did these problems). \n\nThe simple way to address this is to assume that all of the potential energy of the spring .5*k*x^2 is imparted to the freak, and turned into the freak's kinetic energy, .5m*v^2. \n\nAs kirchwey points out, there might very well be a new equilibrium position for the spring due to the freak's weight, or at least the component of the freak's weight that is projected along the axis of the tube. If the spring was not at all compressed by anything but the freak's weight, then if the spring was "released" the freak would not go anywhere. What one has to decide is, is the equilibrium point the problem mentions the equilibrium point for an unloaded spring, or the spring with freak compressing it a bit. Whatever you decide will determine how you solve this problem. I would try it both ways and see what the different answers are (they should not be too different because the freak does not weigh that much, relative to the spring force).
278183
Depends on the weapon, the trajectory, the type of ammunition, etc. But if you shot someone straight up the rectum with something like a .45, they won't have to worry about kissing their arse goodbye, because death will most probably come very quickly.
41449
When an object is launched under the influence of only a uniform gravitational force field, it is said to follow a projectile motion. Under ideal condition, the path it traces is parabolic. There is catch here – if the initial velocity is equal or more than the escape velocity then the object will simply vanish into space i.e. leave earth!\n\nImagine if the launch angle from the horizontal is “theta” and initial velocity is “U” then you can easily compute vertical (Uv) and horizontal (Uh) components of it. The Uv will be equal to U x Sin(theta) and Uh will be equal to U x Cos(theta). Now by using Uv you can compute the max height it will reach and the total time it will take to go up and return to ground. With this round trip time in hand, you can easily compute the max horizontal distance it will cover before hitting the ground.\n\nAll these calculation assume ideal conditions like no wind resistance etc.
37982
Use kinematics! Assuming air resistance to be negligible, as it should be for an arrow, the horizontal velocity of a projectile (i.e. an arrow) is constant. Thus, the velocity at which the arrow leaves the bow is constant after it has left the bow. Time the amount of time it takes for the arrow to reach the target after it left the bow. Then, use the fact that v=x/t to find the horizontal velocity of the arrow. Since if the arrow is fired horizontally from the bow, the only velocity imparted to the arrow from the bow is horizontal, the muzzle velocity should be equal to the horizontal velocity that the arrow has in flight. The vertical velocity of the arrow will be constantly changing through out its flight (it will have a constant downwards accelleration due to gravity of 9.81 m/s^2) but, assuming once more that the arrow is horizontal in the bow, there will be no vertical motion in the bow. I suggest using a video camera to time the amount of time it takes for the bow to reach the target after it has left the bow. That way you can use a frame by frame method to get a really precise time of flight. The distance the arrow travels during that time is really easy - it's just the distance between your bow and the target.
485020
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
113785
My guess is that you die.  Tornados typically have winds up to 250 mph and throw debris for hundreds of yards.  Being within the 50 feet or so to fire a shotgun at it would be suicide.\n\nBut even if you did shoot the gun the tornado is such a powerful force that a gunshot would have no effect on it.  Now, maybe something like one of the fuel air weapons the military uses might have an effect.
346277
Depends, a 9mm, .45, .223, or 7.62x39, or a .50bmg. \n\nThe only one gauranteed to go through would be the .50bmg.\nBut with a few shots (even as little as 3) the .223 or 7.62x39 could easily go through.  \n\nIn lay man's terms, a handgun, no.  A rifle, maybe.