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I have a formula that calculates $R$ with the following formula:

$R = \dfrac{V\cos(α)\Big(V\sin(α) + \sqrt{V\sin(\alpha))^2 + 2gh}\Big)}{g}$

but in this case I need to know $V$ and I already know $R$ I also know the other terms $a, g, h$. How do I rewrite the formula to get $V$?

I with my little math knowledge this is what i came up with:

$V = \dfrac{R}{\cos(\alpha)}\dfrac{\Big(\dfrac{R}{sin(\alpha)} + \sqrt{\big(\frac{R}{ \sin(\alpha)}\big)^2 + 2gh}\Big)}{g}$

but I am pretty sure this is wrong :(

My math is really rusty, so pls explain it in children terms :)

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2 Answers 2

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Well, we only need a few substitutions to reduce the tedium.

So let $\cos \alpha =c,\sin \alpha =s,$ then the first equation becomes $$Rg=Vc(Vs+\sqrt{V^2s^2+2gh}),$$ which transforms to $$\frac{Rg-V^2cs}{Vc}=\sqrt{V^2s^2+2gh}.$$ This then becomes (after squaring and some other apparent simplifications) $$(Rg-V^2cs)^2=V^2c^2(V^2s^2+2gh).$$ Then we make further substitutions: Letting $Rg=A,cs=B,c^22gh=C,$ we obtain now $$(A-V^2B)^2=V^4B^2+V^2C^2,$$ which eventually simplifies to $$A^2=V^2(2AB+C),$$ so that we have $$V=\pm\sqrt{\frac{A^2}{2AB+C}}.$$ Finally, back substitution gives the desired result.

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You can always put long equations like this in Wolfram Alpha, they offer step-by-step solutions:

Link to solution

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  • $\begingroup$ I just saw the step-by-step solutions are behind a paywall now, sry. But at least you have the solution now. $\endgroup$ Mar 21, 2019 at 16:32

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