Hi please help me solving this equation for $\theta$. Thanks
$R = \frac{V_0^2}{g}(b\sin^2{\theta} + \sin{2\theta})$
That R, g, b and $V_0$ are constant
Solve the equation for $\theta$
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$\begingroup$
calculus
trigonometry
projectile-motion
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0Divide both sides by $\sin^2\theta$ or $\cos^2\theta$ – 2017-02-05
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0Expand $\sin2\theta$ and get $\sin\theta(\sin\theta+2\cos\theta)$ – 2017-02-05
1 Answers
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hint :$$R = \frac{V_0^2}{g}(b\sin^2{\theta} + \sin{2\theta})\\b\sin^2{\theta} + \sin{2\theta}=\dfrac{gR}{v_0^2}\\ b\sin^2{\theta} + 2\sin{\theta}\cos{\theta}=\dfrac{gR}{v_0^2}\\$$now divide by $\cos^2 {\theta}$ then you have a quadratic equation by $\tan \theta$ $$b\tan^2{\theta} + 2\tan{\theta}=\dfrac{gR}{v_0^2}(\dfrac{1}{\cos^2 {\theta}})\\ b\tan^2{\theta} + 2\tan{\theta}=\dfrac{gR}{v_0^2}({1+\tan^2 {\theta}})$$ finally you will have $$(b-\dfrac{gR}{v_0^2})\tan^2\theta+2\tan \theta-\dfrac{gR}{v_0^2}=0\ \tan\theta=\dfrac{-2\pm\sqrt{4+4(b-\dfrac{gR}{v_0^2})(\dfrac{gR}{v_0^2})}}{2(b-\dfrac{gR}{v_0^2})}\\\\$$