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> 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?
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