1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook