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Re: [Phys-l] Linear Air Drag



Peter, by linear air drag do you mean f = -b v where f and v are vectors? If so the vector direction for f changes sign at the apex since v changes sign at that point.

Richard Bowman
Bridgewater College
-----Original Message-----
Peter Schoch asked:

I have a quandry, and am hoping someone can clear it up for me...

I just worked a linear air drag problem for 2-D motion to show my
students. The x direction was not a problem. For the y direction I
worked it to rise to a max height, lower than if no drag were present,
and then switching the sign of the drag term I worked the other half
of the vertical motion. I got what appears to be a perfectly
reasonable solution, whose graph is what I expected (lower y and
shorter in x than the vacuum case).

Now, I've just been given a copy of Taylor's "Classical Mechanics" and
asked to teach the course next semester. In that textbook (which
seems very well done) he does linear drag in Chapter 2. As part of it
he does vertical motion, initial v upward, solves the first order, ODE
and states that it is good for the entire motion!

Now, I admit that my solution matches his for the t=0 to highest point
of rise part. However, for the second part of the motion, with the
drag 'switched' my solution differs slightly from his -- and the
graphs of he two differ (but only very slightly). I checked the
errata for the textbook, and none are mentioned for this.

I need some outside brain power. Why is Taylor's solution correct
(that you need not 'switch' the drag term at the apex of the motion)?

Thank you,
Peter Schoch