Re: Ballistics

[Gurth <Gurth@xxxxxxxxx>]

According to Brian McDonald, on Sat, 20 Jul 2002 the word on the street was...

>   About how far up would a 7.62 Nato round go if fired straight up?  and
> about the distance for a 5.56 Nato round?

This is a fairly basic physics problem, I think, provided you don't consider 
air resistance :) If I remember my high school physics right, two formulas 
apply:

	v(t) = v(0) + a * t
and
	s(t) = s(0) + v(0) * t + 1/2 * a * t^2

a = -9.8 m/s^2 (the acceleration due to gravity, which in this case slows 
down the projectile).
v(0) = the weapon's muzzle velocity.
v(t) = 0 (because we're looking at the highest point of the trajectory).

That lets you find the time the round is in the air; assuming 1,000 m/s 
muzzle velocity (about right for 5.56x45 mm), that gives:

	0 = 1000 - 9.8 * t
	t = 1000 / 9.8
	t = 102 seconds

Fill that in in the other formula, together with s(0) being 0 m elevation (so 
I'm basically assuming the firer is in a pit with the muzzle of the weapon at 
ground level :) and you get:

	s(t) = 0 + 1000 * 102 + 1/2 * -9.8 * 102^2
	s(t) = 102000 - 50979.6
	s(t) = 51020 m

Call it 51 km.

But like I said, this is without taking into account friction, so it really 
only applies on a planet with the same gravity as earth but without any 
atmosphere. It's going to be a lot less if friction is considered as well.

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