# Friction proportional to v^2

1. Sep 12, 2006

### Tomsk

Hi, I'm stuck on this problem.
This was fine, it's the next bit I don't get.
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:

$$-\int{\frac{dv}{g+kv^2}}=t$$

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

$$v=\tan{\theta}$$

$$dv=\sec^2{\theta}d\theta$$

$$\Rightarrow -\int{\frac{d\theta}{g\cos^2{\theta}+k\sin^2{\theta}}$$

Now what? Maybe use:

$$\sin^2{\theta}=\frac{1-\cos{2\theta}}{2}$$

$$\cos^2{\theta}=\frac{1+\cos{2\theta}}{2}$$

? It doesn't seem to be helping!

I just noticed:

$$\int{\tan{x}dx}=\ln{\sec{x}}$$

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...)

Last edited: Sep 12, 2006
2. Sep 12, 2006

### SGT

Try the substitution:
$$v=\sqrt\frac{g}{k}\cdot tan \theta$$

3. Sep 12, 2006

### Tomsk

Awesome, thanks very much! I'll hopefully be able to do the rest now.