How Does Air Resistance Affect the Distance of Free Fall?

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SUMMARY

The discussion focuses on the effect of air resistance on the distance a particle falls in a gravitational field. The participant begins with the equation m(dv/dt) = -mg - kmv², leading to a complex integration involving the inverse tangent function. The correct formula for the distance s from initial velocity v0 to final velocity v1 is provided as s = (1/2k) ln((g - kv0²)/(g - kv1²)). The key insight is the application of the relation &ddot;x = v (dv/dx) to simplify the integration process.

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  • Understanding of Newton's second law of motion
  • Familiarity with differential equations
  • Knowledge of integration techniques, particularly involving inverse functions
  • Basic principles of kinematics in physics
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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

[tex] \ddot{x} = v \frac{dv}{dx}[/tex]

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

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