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.eduhttp://usna.edu/Users/physics/mungan/
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