Question Number 8704 by vuckintv last updated on 22/Oct/16
![Solving for A. U(z) = U_b +((2A)/(h+1))(ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ]](https://www.tinkutara.com/question/Q8704.png)
$${Solving}\:{for}\:{A}. \\ $$$${U}\left({z}\right)\:=\:{U}_{{b}} +\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right] \\ $$
Answered by Rasheed Soomro last updated on 22/Oct/16
![U(z) = U_b +((2A)/(h+1))(ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ] ((2A)/(h+1))(ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ]=U(z) − U_b ((2A)/(h+1))=((U(z) − U_b )/((ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ])) A=(((U(z) − U_b )(h+1))/(2(ρ×g×sin(α))^n [H^(n+1) −(H−Z)^(n+1) ]))](https://www.tinkutara.com/question/Q8706.png)
$${U}\left({z}\right)\:=\:{U}_{{b}} +\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right] \\ $$$$\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right]={U}\left({z}\right)\:−\:{U}_{{b}} \\ $$$$\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}=\frac{{U}\left({z}\right)\:−\:{U}_{{b}} }{\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right]} \\ $$$${A}=\frac{\left({U}\left({z}\right)\:−\:{U}_{{b}} \right)\left({h}+\mathrm{1}\right)}{\mathrm{2}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right]} \\ $$