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Re: [Phys-l] Killed by a Falling Bullet? - Myth busters.




At 07:11 PM 1/21/2007, Jack, you wrote:

....Assuming that a bullet dropped from the
>maximum height of a vertically fired bullet is not lethal in all
>circumstance, what is the minimum deviation from exactly vertical that
>would make the bullet lethal in most circumstances.

///

> Jack

At 11:40 AM 1/22/2007, I wrote:
It may be helpful to pursue this point a little further.
I mentioned that a bullet may hit a live target with greater
than terminal velocity or at terminal velocity.
In the former case, it may well be lethal.
In the latter case it will give an unpleasant bruise, unless it hits
some vulnerable point. The one speed is higher than the other,
of course.

For bullets of the usual calibers, there is no free fall at issue
(except perhaps to high school teachers, for whom the vacuum
approximation is helpful?)
But Jack's question is interesting, and in fact easily answered.
What is a plausible minimal angular offset from the vertical
that will allow a given round to hit a target at (much) greater
than terminal velocity? An iterative method answers this
quickly - but I have no time until this evening.
///
Brian Whatcott


Given a little time to contemplate the puzzle, I see I was
making too much of it.
Suppose that the gun elevation for maximal range is around
40 degrees.
Suppose that at this elevation, the final speed of the bullet
corresponds with its terminal velocity - which is plausible.

Then for any distance shorter than this range, there are two
elevations which will make the target: one greater than
40 degrees, and one less than forty degrees.
It is evident that a higher elevation will provide a final speed
no greater than before, but an elevation lower than the
one needed for maximal range provides a bullet with
increased final speed.

There is apparently a suggestion that angles close to vertical
but much more than the elevation needed for maximum range
could provide lethal speed.
My modeling effort does not confirm this.

Perhaps Jack could demonstrate for a bullet of the caliber
in question: 9mm or 0.3 inch nominal?



Brian Whatcott Altus OK Eureka!