MHB Skydiver/Physics Diff.Eq. Problem

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A skydiver weighing 180 lb falls from 5000 ft and opens the parachute after 10 seconds, with air resistance modeled as \(0.75|v|\) before and \(12|v|\) after the parachute opens. The equation of motion is derived from the forces acting on the skydiver, leading to the differential equation \(m \frac{dv}{dt} + 0.75v = mg\). The discussion seeks to find the skydiver's speed at parachute deployment, the distance fallen before opening, and the limiting velocity after the parachute opens. Participants are encouraged to contribute solutions and insights on these calculations. Overall, the thread focuses on applying physics and differential equations to solve the skydiving scenario.
alane1994
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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 :o
 
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alane1994 said:
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 :o
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:
[math]\sum F = m \frac{dv}{dt}[/math]

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

So the equation is:
[math]m \frac{dv}{dt} + 0.75v = mg[/math]

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

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

-Dan
 
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Thank you very very much for that!
 

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