[Phys-L] Re: thrown ball with air drag

[Carl Mungan <mungan@USNA.EDU>]

>  > A ball is thrown vertically upward with initial speed v0. Assume air
> >  drag is proportional to speed squared. Let the speed of the ball just
> >  before it hits the ground be denoted v. Show that 1/v^2 = 1/v0^2 +
> >  1/vT^2 where vT is the terminal speed of the ball.
> >
> >  I feel like there must be some physical significance as to why this
> >  result comes out as a sum of inverse squares. (Doing the problem does
> >  not really shed any light on this interesting feature, at least the
> >  way I did it.) Anyone have any ideas about it?
>
> Intuitively this result makes sense. If v0 is very large that v=vT,
> like for a parachutes. And v=v0 when v0<<vT. Why are you looking for a
> deeper physical significance?

Because the solution comes out in such a simple and neat form. I feel
like there might be an alternative or qualitative way to understand
it. Don't you automatically do this whenever you grind through a long
mathematical problem and get a simple result: say to yourself, "There
must be a more elegant solution"?

> Suppose you multiply each side by 2/m, where m is the mass of the ball.
> This gives 1/E=1/E0 + 1/ET, where each E stands for kinetic energy.
> Ludwik Kowalski

Yes, I noticed that. I also notice that the drag is proportional to
v^2 and the answer came out as inverse squares. But I still can't see
the physical significance. We can say E=E0-Ethermal where Ethermal is
the energy lost to air drag along the path. If we equate this to your
expression, we find that the fraction of mechanical energy lost is
Ethermal/E0 = 1/(1+ET/E0). Yes this has the right limits, but I'm
still looking for a simple way to see any one of these expressions
without having to integrate. Don't you feel that compulsion? Carl
--
Carl E. Mungan, Asst Prof of Physics  410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-5002
mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/
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