Timeline for Finding the angle required to hit a stationary target

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Oct 28, 2020 at 17:06	comment	added	Michael Seifert		@Undergrad2019: You can't "do" anything with it. You only have one equation relating two unknowns ($\theta$ and $x$). So given $x$ you can find $\theta$, or vice versa; but you can't solve for both simultaneously.
Oct 28, 2020 at 14:58	comment	added	Undergrad2019		what do I do with the $x$ which is unknown?
Oct 28, 2020 at 14:56	comment	added	Undergrad2019		$\theta =\arctan\left(\frac{v_0}{xg}\pm \sqrt{\frac{v_0^2\left(v_0^2-2gk\right)}{x^2g^2}-1}\right)$
Oct 28, 2020 at 14:53	comment	added	Undergrad2019		$S_0$ is the $y$ coordiante of our cannon, while $S(t)$ is our $y$ coordinate of the target. I get that $S(t) =-\frac{gx^2}{2v_0^2}\tan^2\theta +x\tan\theta - \frac{gx^2}{2v_0^2}+ S_0 - S(t) = 0$ we can write $S_0-S(t) = K$ for now, letting $u=\tan\theta$ since everything but $\theta$ is constants we have a second order polynomial with $K=m_1u+m_2u^2$ who has the solutions $u= \frac{-a\pm \sqrt{a^2+4kb}}{2b}$ implying that $\theta = \arctan\left(\frac{-a\pm \sqrt{a^2+4kb}}{2b}\right)$
Oct 28, 2020 at 14:20	history	edited	Michael Seifert	CC BY-SA 4.0	added 24 characters in body
Oct 28, 2020 at 13:40	history	answered	Michael Seifert	CC BY-SA 4.0