Rearranging formula for z

**Haris** · November 22nd 2010, 08:39 AM

(z+h)+[(Q/b*(z+h))^2/(2*g)]=z

Could someone please solve for z.

We know all the variables apart from z.

Cheers

**tonio** · November 22nd 2010, 10:18 AM

Originally Posted by Haris

(z+h)+[(Q/b*(z+h))^2/(2*g)]=z

Could someone please solve for z.

We know all the variables apart from z.

Cheers

What you wrote is, apparently, $\displaystyle{z+h+\left[\frac{Q}{b}\cdot\frac{(z+h)^2}{2g}\right]}=z\Longrightarrow \displaystyle{h+\left[\frac{Q}{b}\frac{(z+h)^2}{2g}\right]}=0\Longrightarrow \displaystyle{(z+h)^2=-\frac{2bgh}{Q}}$ ...

Yet the above is so simple that perhaps you meant another first expression...

Tonio

**Haris** · November 22nd 2010, 11:24 AM

$z+h+(\frac{Q}{b(z+h)})^2*\frac{1}{2*g}=z$

sorry, it was meant to look like this.

**pickslides** · November 22nd 2010, 11:46 AM

$\displaystyle z+h+\left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{ 2g}=z$

Subtract $z$ from both sides

$\displaystyle h+\left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{2g }=0$

Take $h$ from both sides

$\displaystyle \left(\frac{Q}{b(z+h)}\right)^2\times\frac{1}{2g}=-h$

Now lets simplify this a bit

$\displaystyle \frac{Q^2}{b^2(z+h)^2}\times\frac{1}{2g}=-h$

$\displaystyle \frac{Q^2}{2gb^2(z+h)^2}=-h$

Now your turn, multiply both sides through by $\displaystyle (z+h)^2$

**Wilmer** · November 22nd 2010, 12:07 PM

Well, a "z" cancels out from each side (why even bother showing those?!),
leaving a less wieldy:
Q^2 / [b(z + h)]^2 = -2gh
Surely you can handle that...

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