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Wierd equation

  1. May 28, 2005 #1
    Guys, I'm a little confused. Please, take a look at this:

    [tex]L=\frac{u\cos A}{r} \left( 1 - e^{-rt} \right)[/tex]

    Solving for [tex]t[/tex] gives

    [tex]t=\frac{1}{r}\ln \left( \frac{u\cos A}{u\cos A - rL} \right)[/tex]

    Then, substitute [tex]t[/tex] in the equation that follows

    [tex]H=-\frac{gt}{r}+ \frac{1}{r}\left( u\sin A + \frac{g}{r} \right) \left( 1 - e^{-rt} \right) + h[/tex]

    which gives

    [tex]H = \frac{g}{r}\ln \left( 1 - \frac{rL}{u\cos A} \right) + \frac{L}{u\cos A} \left( u\sin A + \frac{g}{r} \right) + h[/tex]

    Let's say we're given the values:






    That implies we haven't yet obtained [tex]u[/tex] and [tex]A[/tex]. There is ONE equation and TWO variables. However, we're looking for the [tex]u_{min}[/tex] AND [tex]A_{max}[/tex]. I think we first need to solve the equation above for [tex]u[/tex], take the first derivative of the expression with the variable [tex]A[/tex], set it equal to zero, and then solve it for [tex]A_{max}[/tex]. Consequently, we're are able to get [tex]u_{min}[/tex].

    I've had difficulty solving the equation for [tex]u[/tex]. I also tried to solving it with aid of the computer, but it won't give me the answer!!!

    Any help is highly appreciated.
  2. jcsd
  3. May 28, 2005 #2


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    Science Advisor
    Homework Helper

    First, check your substitution. It does not look right to me. Either that or your H equation is wrong to begin with.
    Last edited: May 29, 2005
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