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He asked about what would happen to a parachutist falling with a large rockRichard M. Langer Gateway High School,
attached to him from a string. He (and the rock) would reach a terminal
speed (he reasoned) of some value lets call it V. Then the parachutist
released the rock. He reasoned that he should slow down (ie + upward
acceleration) because his terminal speed without the rock was some value v,
which is smaller than V. No arguements with his reasoning.
At first I was confused by this as I had just told them that for a falling
mass m, the air resistance force would be build to a value of mg and then
remain at that value and therefore never be greater than mg.
So we did it on the fly in the classroom, but not for the release of the
rock. I sat here this am and did the "experiment" in IP. Of course he was
right, but only after doing it in IP was I able to see exactly why (so I
could explain it to them!!).
Once the system (rock plus parachutist) reaches terminal velocity, the NET
force of air resistance is equal to the NET weight. BUT, the force of air
resistance on the parachutist is GREATER than his weight and the force of
air resistance on the rock is LESS THAN its wieght. (net force on parachute
and rock individually are balanced to zero by the tension in the string)
Therefore when the rock is released, there is a net UPWARD force on the
parachutist which slows him down, eventually he reaches his terminal speed
v. Do you know then what happens to the rock?
If anyone has IP 2.5 and wants a copy of this file to play with email me
and I will email it back as an attachment.
forever learning new ways to look at things!
Cindy Schwarz
Associate Professor of Physics
Vassar College
Poughkeepsie
P.S. Now I am going to keep IP handy in my class to look at some of these
things that come up on the spot!!