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Skydiver/Physics Diff.Eq. Problem

alane1994

Active member
Oct 16, 2012
126
A skydiver weighing 180lb (including equipment) falls vertically downward from an altitude of 5000 ft and opens the parachute after 10 s of free fall. Assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude \(0.75|v|\) when the parachute is closed and is of magnitude of \(12|v|\) when the parachute is open, where the velocity \(v\) is measured in ft/s.

a) Find the speed of the skydiver when the parachute opens.
b) Find the distance fallen before the parachute opens.
c) What is the limiting velocity \(v_L\) after the parachute opens?

Any and all assistance would be appreciated greatly. And thanks for putting up with all of my questions :eek:
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
A skydiver weighing 180lb (including equipment) falls vertically downward from an altitude of 5000 ft and opens the parachute after 10 s of free fall. Assume that the force of air resistance, which is directed opposite to the velocity, is of magnitude \(0.75|v|\) when the parachute is closed and is of magnitude of \(12|v|\) when the parachute is open, where the velocity \(v\) is measured in ft/s.

a) Find the speed of the skydiver when the parachute opens.
b) Find the distance fallen before the parachute opens.
c) What is the limiting velocity \(v_L\) after the parachute opens?

Any and all assistance would be appreciated greatly. And thanks for putting up with all of my questions :eek:
To set this up I will assume that we have a y coordinate system such that y = 0 ft at the point when the skydiver leaves the plane, so v = 0 ft/s both at t = 0 s, and I am setting the positive y direction to be downward.

As the diver is falling we have the following derivation of the equation of motion:
\(\displaystyle \sum F = m \frac{dv}{dt}\)

\(\displaystyle mg - 0.75v = m \frac{dv}{dt}\)
(mg is the weight and the friction force 0.75v points upward, the negative direction.)

So the equation is:
\(\displaystyle m \frac{dv}{dt} + 0.75v = mg\)

And of course distance is the time integral of the speed....

See what you can do with this and let us know.

-Dan
 
Last edited:

alane1994

Active member
Oct 16, 2012
126
Thank you very very much for that!!