```Date: Jul 29, 1997 11:43 AM
Author: tony richards
Subject: Re: Flying projectile eqn

dX/dt = V*cos(angle) = instantaneous horizontal velocity = VxdY/dt = V*sin(angle)-g*t = instantaneous vertical velocity = VyV= muzzle velocity (velocity of bullet leaving gun)angle = angle to horizontal of gun barrelg = local acceleration due to gravity = 9.81 metre/sec/sec at earth's surface(if gun is on another planet, g will be the value for that planet of course)t = time after bullet leaves muzzle of gunso at any instant t after the bullet leaves the muzzleX = Xgun + V*cos(angle)*tY = Ygun + V*sin(angle)*t - g*t^2Xgun, Ygun = co-ordinates of end of gun barrel when bullet leaves it(assumed given).At any time t after leaving the gun, the velocity of the bullet, Vbullet isVbullet=sqrt(Vx^2+Vy^2) at an angle angbull to the horizontal, wheretan(angbull)= Vy/Vx.The above ignores the effects of air resistance and the curvature and rotationof the earth, effects which matter and have to be taken intoaccount for accurate artillery fire over long distances.-- Tony Richards            'I think, therefore I am confused'Rutherford Appleton Lab  'UK                       '
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