more inegration

**billym** · March 25th 2008, 11:12 PM

I need to solve for v... not a clue...

$ \frac{dv}{dt} = -\frac{g}{b^2}(v^2 + b^2) $

**TheEmptySet** · March 25th 2008, 11:25 PM

Originally Posted by billym

I need to solve for v... not a clue...

$ \frac{dv}{dt} = -\frac{g}{b^2}(v^2 + b^2) $

This is a seperable equation

$ \frac{dv}{v^2+b^2} = -\frac{g}{b^2}dt $

Now we integrate both sides

$ \int \frac{dv}{v^2+b^2} = \int -\frac{g}{b^2}dt \iff \frac{1}{b^2} \int \frac{dv}{1+(v/b)^2}= - \frac{g}{b^2} \int dt \ =$

$\int \frac{dv}{1+(v/b)^2} = -g \int dt $
finally we take the integral and we get

$b \tan^{-1}(\frac{v}{b})=-gt+c$

Now all you need to do is solve for v.

Good luck

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