- #1

DivGradCurl

- 372

- 0

[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:

[tex]g=32[/tex]

[tex]h=3[/tex]

[tex]r=\frac{1}{5}[/tex]

[tex]L=350[/tex]

[tex]H=10[/tex]

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.