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Topic: Flying projectile eqn
Replies: 2   Last Post: Jul 29, 1997 11:43 AM

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 tony richards Posts: 164 Registered: 12/8/04
Re: Flying projectile eqn
Posted: Jul 29, 1997 11:43 AM

dX/dt = V*cos(angle) = instantaneous horizontal velocity = Vx

dY/dt = V*sin(angle)-g*t = instantaneous vertical velocity = Vy

V= muzzle velocity (velocity of bullet leaving gun)
angle = angle to horizontal of gun barrel
g = 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 gun
so at any instant t after the bullet leaves the muzzle

X = Xgun + V*cos(angle)*t
Y = Ygun + V*sin(angle)*t - g*t^2

Xgun, 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 is
Vbullet=sqrt(Vx^2+Vy^2) at an angle angbull to the horizontal, where
tan(angbull)= Vy/Vx.

The above ignores the effects of air resistance and the curvature and rotation
of the earth, effects which matter and have to be taken into
account for accurate artillery fire over long distances.

--
Tony Richards 'I think, therefore I am confused'
Rutherford Appleton Lab '
UK '

Date Subject Author
7/27/97 milllen@bbs.maloca.com
7/27/97 William L. Bahn
7/29/97 tony richards