(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

"A passenger (mass m) initially at rest steps out of an airplane. Assume down is the positive x-axis and put the origin at the airplane. Assume the air resistance force is linear in the velocity so F(air)= -mbv. Find the distance D he has fallen when his velocity is v."

2. Relevant equations

Equations of motion

The "vdv/dx trick": d^{2}z/dt^{2}= (z-dot)*(d(z-dot)/dz)

F(tot) = ma

F(air) = -mbv

Weight = mg

3. The attempt at a solution

Here's how far I've gotten:

Since the skydiver is only falling in the x direction, there's only one equation of motion, which I found to be ma = -mbv + mg [or, alternatively, m(x-double dot) = -mb(x-dot) + mg]. Now, I know I want the relation of distance and velocity, without time, so I use the "vdv/dx" trick (so that there's no longer time in the equation).

That makes this mv*(dv/dx) = -mbv + mg, or m(x-dot)*(d(x-dot)/dx) = -mb(x-dot) + mg. I rearranged this to get dx = (-m/b)*(vdx/v-(mg/b)), where -(mg/b) is the terminal velocity.

I'm sorry for all the writing, but am I correct so far? And how do I continue to solve the problem from here? Any help would be appreciated.

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# Classical mechanics - finding distance D in terms of velocity

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