Question 200557
The 8-in(diameter) Howitzer on the U.S. Army's M110 can propel a projectile a distance of 18,500yd. If the angle of elevation of the barrel is 45 degrees, then what muzzle velocity (in feet per second) is required to achieve this distance?
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Max distance (at 45 degs) = v^2/g (this can be derived, if you're interested)
d = 55500 feet
v^2 = 55500*32 = 1776000
v = ~ 1332.66 ft/sec
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The shell diameter doesn't matter, nor the weight of the projectile.
This is assuming no air friction, as usual.  Physics occurs in a frictionless vacuum.