A person riding on the roof of a freight train fires a gun pointed forward...?
962535
a-faster\nb-no difference
350753
wather train?????\ninteresting
916863
I get that the maximum distance that the ship can fire is 6371m, meaning that any distance greater than 3571m from the western shore should be safe.  It may be tempting to assume that the target ship would be safe within a certain minimum distance, where it would be protected by the mountain peak, but the enemy ship could hit any target within the 3571m limit by angling its guns higher than 45 degrees.\n\nEDIT:\nTo start solving our problem, let's figure out the maximum distance that the enemy ship can fire.  It's not hard, and we know we'll have to figure it out anyway.\n\nIf the ship fires its projectile at exactly 45 degrees, the projectile will make a parabolic arc through the air.  At angles of more than 45 degrees, the projectile will fly higher, but it won't go as far (like a pop-up fly in baseball).  At angles less than 45 degrees, the projectile will fly lower, and hit the water with more horizontal motion (like a line-drive in baseball).  At exactly 45 degrees, it will achieve its maximum horizontal distance.  \n\nIn order to figure out how far the projectile flies, we need to break up its motion into horizontal and vertical components.  Imagine a right triangle with a hypotenuse of 250 units and one angle of 45 degrees.  (The other angle will also be 45 degrees, but that doesn't matter for now.)  The line adjacent to our angle will be as long as the projectile's horizontal momentum.  The line opposite the angle will be as long as the projectile's vertical momentum.  We just need a little bit of trigonometry to figure out these numbers.\n\nsin(a) = opposite / hypotenuse;\nsin(45) = Vv / 250;   (Vv represents the vertical velocity)\nVv = sin(45) * 250 = 177m/s;\n\nSince we're using a right triangle with two 45 degree angles, both sides will be the same length.  If we were using a different angle, we could easily solve for the horizontal momentum using the Pythagorean Theorum.\n\nSo, we know that our projectile had a horizontal velocity of 177m/s and a vertical velocity of 177m/s.  Let's figure out how long it would stay in the air.  We know that it's constantly accelerating towards the earth due to the force of gravity.  Our answer lies in the definition of acceleration: acceleration is a change in velocity over time.  Mathematically, that's (a=ΔV / t).  The "Δ" symbol is the Greek letter delta, often used to represent a change in a value.  \n\nTo figure out how long it stayed up, we only need to worry about the projectile's vertical velocity.  We know that it's initial vertical velocity is 177m/s.  We also know that it's vertical velocity at the top of the arc is 0m/s.  The change between these two points in time is (177m/s - 0m/s =) 177m/s.  The only other thing we need to figure out time is the acceleration due to gravity.  On Earth, it's always 9.81m/s^2. So...\n\na = ΔV / t;\n9.81 = 177 / t;\nt = 177 / 9.81 = 18.04 seconds.  That's how long it takes the projectile to reach the top of it's arc.  It will take equally as long to reach the bottom of it's arc, so the total flight time is 36.09 seconds.  \n\nNow we have a simple distance equation.  Since distance is equal to velocity multiplied by time, the projectile's horizontal distance will be equal to its horizontal velocity multiplied by the amount of time that it stayed in the air.\n\nd = 177 * 36 = 6372m.  (This number is a little higher than the first one, because I rounded a little differently.  It doesn't make a huge difference.)  If we take out the distance to the western shore (2500m + 300M), we get a maximum distance of 3572m from the western shore.  Any ship beyond that distance will be safe.  \n\nAll we need to do now is figure out whether or not the enemy ship can fire at that maximum distance.  If the mountain is too high, the ship will have to increase the angle of its guns, thereby losing range.  We can figure out how high our projectile is going to fly with
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.
1093046
Jumping the Gun a bit !!!!
565420
A BULLET if you get caught. LOL
604300
A blonde at a blinking railroad light??
25818
The gun relies on a compressed air cartidge.  If the cartridge is removed, the taser will not fire.  I hope this helps.
639219
it was fired. lots of times. \nstationed firmly or mounted on a railway wagon\n\ntry this http://www.waffenhq.de/panzer/dickeberta.html\n\ntranslation of text would be difficult (much technical), but nice pic. click an "bildergalerie" below
850728
he received a shot gun????\n\ndats very stupid!!! and bad influence for the kids.
158383
You have to shoot it from a far enough distance not to make it go farther than the half meter the bullet would cause it to go.It would also depend on where you hit the tin can. That would have an effect on it's travel.
718712
that would depend on the firing speed.
1075044
There is a problem with the cocking mechanism.  appears that the sear, (which holds the bolt back) isn't engaging and that is why it keeps firing.  I can't tell you more without actually seeing inside your gun.
534562
depends on which bullet hit him ,not through the back, but the one that went through Governor O'Connelly's leg, ricocheted back behind the limo and made a u-turn into JFK's skull. Eye witness testimony have 2 women who said they saw a railroad-uniformed worker put a rifle case on the RR control box next to the RR tracks just behind the hedges on the famous grassy knoll and watched a man in a suite they thought was secret service assemble the rifle and shoot over the hedges and put the gun back in the case at which the man in the RR uniformed removed his coveralls and walked away with the gun. Much speculation. LBJ (V.P.) was known for his bitter contempt for JFK and was quoted at a party the night before that we wouldn't have to deal with that S.O.B after tomorrow. The Warren commission states a single gunman but eye wittiness' state otherwise. The History channel did a week long special in 2003 ( the 40th anniversary of his assassination ) and mater of factly said there were two or three gunmen. I think most people think it was an inside job.
95569
-depends on the shooter!!
491324
With six shots from a gun.
817224
This is for a novel and you came to this forum for research?\n\nNot a good idea.\n\nSeriously though, the rifle issued would be an M-16. he would\nalso carry a Beretta 9mm sidearm.\n\nThank you very much, while you're up.