How Does Air Resistance Affect the Distance of Free Fall?

AI Thread Summary
The discussion focuses on calculating the distance a particle falls under the influence of gravity and air resistance, modeled as a force proportional to the square of the velocity. The user starts with the equation of motion m(dv/dt) = -mg - kmv² and attempts to integrate it, leading to a complex function involving the inverse tangent. They seek clarification on whether a second integration is necessary to find the distance fallen. A suggestion is made to use the relation \ddot{x} = v(dv/dx) to simplify the problem and achieve the correct distance formula. The conversation emphasizes the importance of correctly applying integration techniques in physics problems involving resistance.
eku_girl83
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Here's the problem:
Consider a particle of mass m whose motion starts from rest in a constant gravitational field. If a resisting force proportional to the square of the velocity (kmv2) is encountered, find the distance s the particle falls in accelerating from v0 to v1.

I began with the equation m (dv/dt)=-mg-kmv2.
From this, we see that dv = -g-kv2 dt.
When I separate variables and integrate this however, I get a function that involves the inverse tangent. I would have to integrate a second time to find distance, correct?
The book gives the distance s from v0 to v1 as 1/2k ln ((g-kv02)/(g-kv12)). What am I doing wrong?

Any help greatly appreciated!
 
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Use the relation

<br /> \ddot{x} = v \frac{dv}{dx}<br />

on the left hand side of your equation and you should find things work out :)
 
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