Basic rearranging

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• Apr 7th 2010, 01:35 AM
jimss2
Basic rearranging
I am currently solving questions preparing for my mechanics exam and I am struggling with the final part of the question I have the following equation

y-yo=Vo*sin(beta)*x^2/Vo*cos(beta)-1/2*g*x^2/Vo^2cos^2(beta)

I need to solve this for Vo any help would be much appreciated
• Apr 7th 2010, 02:05 AM
Bacterius
I'm sorry but this is ambigious. I'll just put what I think the equation is into proper LaTeX form, and please tell us if this is the right one.

$\displaystyle y - y_0=V_0 \times \sin{(\beta)} \times \frac{x^2}{V_0} \times \cos{(\beta)} - \frac{1}{2} \times g \times \frac{x^2}{V_0^2} \times \cos^2{(\beta)}$

Is this the one ? (Whew)
• Apr 7th 2010, 02:31 AM
jimss2
Sorry your right that is the correct equation could you please point me in the right direction on how to solve for Vo
• Apr 7th 2010, 02:31 AM
HallsofIvy
Assuming that is correct the $\displaystyle V_0$ and $\displaystyle \frac{1}{V_0}$ in the first term will cancel leaving
$\displaystyle y- y_0= sin(\beta)cos(\beta)-\frac{1}{2}g\frac{x^2}{V_0^2}cos^2(\beta)$

Subtract $\displaystyle sin(\beta)cos(\beta)$ from both sides:
$\displaystyle y- y_0- sin(\beta)cos(\beta)= -\frac{1}{2}g\frac{x^2}{V_0^2}cos^2(\beta)$

Multiply both sides by $\displaystyle V_0^2$:
$\displaystyle V_0^2(y- y_0- sin(\beta)cos(/beta))= -\frac{1}{2}gx^2cos^2(\beta)$

Divide both sides by $\displaystyle y- y_0- sin(\beta)cos(/beta)$:
$\displaystyle V_0^2= -\frac{gx^2cos^2(\beta)}{2(y- y_0- sin(\beta)cos(\beta)}$

And, finally, take the square root of both sides:
$\displaystyle V_0= \sqrt{-\frac{gx^2cos^2(\beta)}{2(y- y_0- sin(\beta)cos(\beta))}}$

$\displaystyle V_0= \sqrt{\frac{gx^2 cos^2(\beta)}{2(sin(\beta)cos(\beta)- y+ y_0)}}$
• Apr 7th 2010, 05:05 AM
jimss2
thankyou this has been very helpful