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Friction proportional to v^2

  1. Sep 12, 2006 #1
    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:

    [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...)
     
    Last edited: Sep 12, 2006
  2. jcsd
  3. Sep 12, 2006 #2

    SGT

    User Avatar

    Try the substitution:
    [tex]v=\sqrt\frac{g}{k}\cdot tan \theta[/tex]
     
  4. Sep 12, 2006 #3
    Awesome, thanks very much! I'll hopefully be able to do the rest now.
     
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