Question Number 8704 by vuckintv last updated on 22/Oct/16
$${Solving}\:{for}\:{A}. \\ $$$${U}\left({z}\right)\:=\:{U}_{{b}} +\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right] \\ $$
Answered by Rasheed Soomro last updated on 22/Oct/16
$${U}\left({z}\right)\:=\:{U}_{{b}} +\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right] \\ $$$$\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right]={U}\left({z}\right)\:−\:{U}_{{b}} \\ $$$$\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}=\frac{{U}\left({z}\right)\:−\:{U}_{{b}} }{\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right]} \\ $$$${A}=\frac{\left({U}\left({z}\right)\:−\:{U}_{{b}} \right)\left({h}+\mathrm{1}\right)}{\mathrm{2}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right]} \\ $$