# Wierd equation

1. May 28, 2005

Guys, I'm a little confused. Please, take a look at this:

$$L=\frac{u\cos A}{r} \left( 1 - e^{-rt} \right)$$

Solving for $$t$$ gives

$$t=\frac{1}{r}\ln \left( \frac{u\cos A}{u\cos A - rL} \right)$$

Then, substitute $$t$$ in the equation that follows

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

which gives

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

Let's say we're given the values:

$$g=32$$

$$h=3$$

$$r=\frac{1}{5}$$

$$L=350$$

$$H=10$$

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

I've had difficulty solving the equation for $$u$$. 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. May 28, 2005

### OlderDan

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