Projectile with a linear air resistance force question

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SUMMARY

The discussion focuses on deriving the maximum horizontal range of a projectile subjected to linear air resistance, represented by the force Fs = −mkv. The equation governing this range is (v0 + g/k)(x/u0) + (g/k^2)ln(1 - kx/u0) = 0, where u0 = V cos(theta) and v0 = V sin(theta). The solution requires numerical methods, such as MAPLE, or approximations for small k values, leading to the formula xmax = 2v0u0/g - 8v0*v0*u0*k/3g^2 for maximum range.

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Homework Statement



OK I hope someone will help me with this

For a projectile with a linear air resistance force Fs =−mkv, show that the maximum horizontal range is given by the equation

(v0 + g/k)(x/u0) + (g/k^2)ln(1 - kx/u0) = 0


where u0 = V costheta, v0 = V sintheta are the horizontal and vertical components of the initial velocity.

The above equation cannot be solved in closed form, either it is solved numerically (e.g. using MAPLE) or it may be approximated, assuming that the correction with k =/= 0 is small. For the latter method show that

xmax = 2v0u0/g - 8v0*v0*u0*k/3g^2


Homework Equations


(v0 + g/k)(x/u0) + (g/k^2)ln(1 - kx/u0) = 0


The Attempt at a Solution



I've done the first part (finding the equation given) but I really don't know what the second part means (the bit about approximation). I tried using taylor series to 2 terms but this just gives the first term not the second, so I'm stumped. Can anyone shed some light on this?
 
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I'm going to assume that the equations given are:

[tex] \frac{x}{v_{0x}} \left ( v_{0y} + \frac{g}{k} \right ) + \frac{g}{k^2} {\ln \left ( 1 - \frac{kx}{v_{0x}} \right ) } = 0[/tex]

and

[tex] x = \frac{2v_{0y}v_{0x}}{g} - \frac{8k v_{0y}v_{0x}}{3g^2}.[/tex]

Besides that, I have absolutely no idea. >_<
 

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