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\begin{equation} -15=40\tan(53) - 4.9\left(\frac{40}{v\cos(53)}\right)^2 \end{equation}

My Question:

What steps should I take algebraically to solve for positive $v$ neatly? Is the only way to solve this by getting it in quadratic form? If so, how can I do that without making it look too messy?

I found this equation by manipulating the kinematics equations in my physics course to solve for one of the variables $v_0$ in a particular problem.

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up vote 2 down vote accepted

Don't let the trig functions fool you-those are just constants. Your equation is $a=b-\frac c{v^2}$, so you can take this to $\frac c{v^2}=b-a$, invert both sides, multiply by $c$ and take the square root, remembering the $\pm$ sign. No need for the quadratic formula.

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Thank you, that advice was roughly 10 times more effective than what the T.A's tried to do. – Leonardo Oct 16 '12 at 23:21

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