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

[rlamont <rlamont@POSTOFFICE.PROVIDENCE.EDU>]

I think Michael has the correct analogy. The resistive force, R,
can be written as b*v^2  in this case. Rewrite the final result
as

1/(b*v^2) = 1/(b*vo^2) + 1/(b*vt^2)  and this becomes

1/R  = 1/Ro + 1/Rt

which is an exact analogy to parallel resistors.

Bob at PC

> -----Original Message-----
> From: Forum for Physics Educators [mailto:PHYS-
> L@list1.ucc.nau.edu] On Behalf Of Michael Burns-Kaurin
> Sent: Thursday, September 15, 2005 3:17 PM
> To: PHYS-L@LISTS.NAU.EDU
> Subject: Re: thrown ball with air drag
>
> It was previously pointed out that the result could be viewed
as
> involving
> inverses of kinetic energy.  In that same vein, the inverses
are of
> heights
> that the ball could reach if it had a certain speed, absent
drag.  I can't
> make sense of that either, however.
>
> Sometimes an equation involving inverses tells us that we are
focusing
> on
> the wrong idea--for instance, resistors in parallel lead to an
equation
> involving inverses, because a more useful quantity is
conductance in
> that
> case.  I don't know how to fit it in, but perhaps time is a
useful quantity
> to look at.
>
> Michael Burns-Kaurin
> Spelman College
>
>
>
>
>
>                       Carl Mungan
>                       <mungan@USNA.EDU>        To:       PHYS-
> L@LISTS.NAU.EDU
>                       Sent by: Forum           cc:
>                       for Physics              Subject:  thrown
ball with air drag
>                       Educators
>                       <PHYS-L@list1.ucc
>                       .nau.edu>
>
>
>                       09/14/2005 03:17
>                       PM
>                       Please respond to
>                       Forum for Physics
>                       Educators
>
>
>
>
>
>
> Someone just came in and asked me a nifty problem. After
diddling
> around a little we got the answer, but you all might enjoy it
too:
> 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? 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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