Hi, I'm stuck on this problem.(adsbygoogle = window.adsbygoogle || []).push({});

This was fine, it's the next bit I don't get. A particle moves vertically under gravity and a retarding force proportional to the square of the velocity. If v is its upwards or downwards speed, show that

[tex]\dot{v}=\mp g-kv^2[/tex] (1)

Respectively, where k is a constant

From (1) I'm thinking I'll integrate up to get v, then again to get z, as v=dz/dt. The other alternative would be to treat is as a second order DE for z, but I tried that and didn't get anywhere. I also tried an integrating factor, but that didn't work either. I've currently got: If the particle moves upwards, show that it's position at time t is given by

[tex]z=z_{0} + \frac{1}{k}\ln{\cos{[\sqrt{gk}(t_0-t)]}[/tex]

Where z_0 and t_0 are integration constants. You may find the identity

[tex]\ln{\cos{x}}\equiv\frac{-1}{2}\ln{(1+\tan^2{x})}[/tex]

useful.

[tex]-\int{\frac{dv}{g+kv^2}}=t[/tex]

But I can't integrate it! I tried the sub

[tex]v=\tan{\theta}[/tex]

[tex]dv=\sec^2{\theta}d\theta[/tex]

[tex]\Rightarrow -\int{\frac{d\theta}{g\cos^2{\theta}+k\sin^2{\theta}}[/tex]

Now what? Maybe use:

[tex]\sin^2{\theta}=\frac{1-\cos{2\theta}}{2}[/tex]

[tex]\cos^2{\theta}=\frac{1+\cos{2\theta}}{2}[/tex]

? It doesn't seem to be helping!

I just noticed:

[tex]\int{\tan{x}dx}=\ln{\sec{x}}[/tex]

So I think I must want v=tan(f(t)), or something. But I just don't see how I can get it. Maybe I have the wrong sub as I want sec^2 to integrate to tan.

Thanks in advance! (Let's hope this tex works...)

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# Friction proportional to v^2

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