Physics question

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ups.com...

You're still assuming he's leading the ME-109 as if the gunner
were in a fixed location. He's assuming the gunner is actually
trying to hit the plane, and smart enough to figure out how
so that he times his shot *correctly*. You're both assuming
he fires straight back. He can hit the Me-109 by firing straight
back, but ONLY if he fires exactly as B-17 crosses the flight
path of the Me-109, and only if the Me-109 is fast enough or
close enough to get there before the bullet drops out of the
flight path of the Me-109. If the Me-109 is at a range of 50
yards, and no slower than the bullet the bullet will be no
more than 7 inches lower than the muzzle of the gun and will
still hit the Me-109 assuming the two planes are at the same
altitude.

Fred...............

I'll say it again - you misunderstand the question. I have stated it as
clearly as I am able and even drawn a diagram. I don't know how else to
explain it. I'm a pilot. I talk with my hands (hopefully without resorting
to certain gestures).

The B-17 is flying West. The ME-109 is flying South. They are both at the
same altitude. There is no wind. The ME-109 will cross the flight path of
the B-17 and the B-17 will, at that moment, be fifty yards from the ME-109.
The tail gunner fires his weapon straight behind the B-17.

There is no way that he can hit the ME-109, given the stated speed of the
B-17 and the muzzle velocity of the bullet.

The ME-109 will *always* be fifty yards or more away from the bullet.
Always.

Rich S.