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

[Jack Uretsky <jlu@hep.anl.gov>]

I think you are in error in your use of terminal velocity (note that I am 
able to say this without impugning your intelligence, morality, or taste 
in fine dining).  Terminal velocity usually refers to the velocity at 
which the "force of gravity", namely mg, is equal to the retarding force 
(ordinarily viscous) of air.  I t refers only to the limiting velocity in
free fall.  A meteorite, for example, entering that atmosphere at a 
supersoning velocity, may remain supersonic until it hits the earth.
Correspondingly, a bullet fired upward at an angle to the vertical may 
have a velocity exceeding "terminal" by an amount sufficient to exceed terminal 
until it strikes the earth and the unlucky child in the flight path.
						Regards,
							Jack


On Mon, 22 Jan 2007, Brian Whatcott wrote:

> > 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!
>
>
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