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The acceleration due to gravity at earth's surface is approximately 9.81 m/s^2. Also note that when a bullet is fired horizontally from a gun is has zero velocity in the y (up and down) direction to start with. Thus it's velocity in the y direction v = a*t where t is time. Taking the integral of time gives us an expression for distance fallen d = 1/2*a*t^2 (assuming zero initial velocity in y direction. This can be rearranged to solve for the time taken to fall to the ground. t = square.root(2*d/a)
So if we assume that the rifle is held 1.5 meters off the ground (approx 5 ft) then from the time the bullets leaves the muzzle it will be approximately .55 seconds until it falls to the ground. So if the bullet has an average velocity of 3000 ft/s (I don't know how much velocity a bullet normally losses in flight so this number may need to be tweaked) if would cover 1650 ft or just under a third of a mile.
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This Board is great
Do you have a slide rule? I do!
WildwritethatformulaoutAlaska
PS...I cant help myself...what would ESCAPE velocity of a 55 grain projectile need to be? Is that a trick question?