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

[Ludwik Kowalski <kowalskil@MAIL.MONTCLAIR.EDU>]

On Thursday, Sep 15, 2005, at 06:03 US/Pacific, Carl Mungan wrote:

> > > 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/

You can say that

1/v^2 = 1/v0^2 +1/vT^2

can be graphically represented by a right triangle. But the same is
true for the thin lens formula or for two parallel resistors. It is not
at all obvious that such common property is indication of something
significant.

Ludwik Kowalski
Let the perfect not be the enemy of the good.
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