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Ignoring air resistace, I have that the height of our canon is $8.2$m and the height of the target is $6.34$m. And that our initial veocity $v_0=18$m/s. I have to determine a way to find the angle required to hit this target.

My work: We have our initial position $(0, 8.2)$ and target is at $(x,6.4)$

The component of the initial velocities would be $$v_{0x}=v_0\cos\theta\\v_{0y}=v_0\sin\theta$$

we also have that $$x=x_0+v_{0x}t\\ y = y_0+v_{0y}t+\frac{1}{2}a_yt^2$$

Where $a_y=-g$ so we have a height function $$S(t) = S_0+v_{0y}t-\frac{1}{2}gt^2$$. I also know that the range is given by $R=\frac{v_0^2\sin 2\theta}{g}$ We can find an expression for $t$ from the $x$ position equation which gives us $t=\frac{x}{v_0\cos\theta}$ Knowng all of this we see that $$S(t) = S_0+x\tan\theta -\frac{gx^2}{2v_0^2\cos^2\theta}$$ we have two unknowns now, namely $x,\theta$. So since the range is given by $R=\frac{v_0^2\sin 2\theta}{g}$ our target will be in $R$ so we have that $x=R$, then we get

$$S(t)=S_0+\left(\frac{v_0^2\sin 2\theta}{g}\right)\left(\frac{\sin\theta}{\cos\theta}\right)-\left(\frac{g}{2v_0^2\cos^2\theta}\right)\left(\frac{v_0^2\sin 2\theta}{g}\right)^2\\= S_0+\left(\frac{2v_0^2\sin \theta\cos\theta}{g}\right)\left(\frac{\sin\theta}{\cos\theta}\right)-\left(\frac{g}{2v_0^2\cos^2\theta}\right)\left(\frac{2v_0^2\sin \theta\cos\theta}{g}\right)^2$$

but now we see that they cancel eachother out, this is where I don't know how to continue, is it wrong to say that $x=R$? I need to know what $x$ is in order to find $\theta$. It also says in the problem that we need to make sure that the target is reasonably far away from the cannon

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The "range equation" only applies on level ground, when the target and the cannon are at the same height. So it is indeed wrong to say that $R = v_0^2 \sin 2 \theta / g$. This is a common error among intro physics students; it's a nice, convenient, compact, easy-to-understand, and useful result, and it's really tempting to use it at every opportunity. But you always have to remember the assumptions that went into it.

Your height equation, $$S(t) = S_0+x\tan\theta -\frac{gx^2}{2v_0^2\cos^2\theta},$$ is in fact correct (since you didn't use the range equation to get it). This provides a relationship between $x$ and $\theta$, since you know $S_0$ and $S(t)$ (what are they?) Actually solving this equation to find $\theta$ as a function of $x$ is trickier, and is more a matter of algebra than of physics. As a hint, try applying the identity $\sec^2 \theta = 1 + \tan^2 \theta$.

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  • $\begingroup$ $S_0$ 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 $\theta = \arctan\left(\frac{-a\pm \sqrt{a^2+4kb}}{2b}\right)$ $\endgroup$
    – Undergrad2019
    Commented Oct 28, 2020 at 14:53
  • $\begingroup$ $\theta =\arctan\left(\frac{v_0}{xg}\pm \sqrt{\frac{v_0^2\left(v_0^2-2gk\right)}{x^2g^2}-1}\right)$ $\endgroup$
    – Undergrad2019
    Commented Oct 28, 2020 at 14:56
  • $\begingroup$ what do I do with the $x$ which is unknown? $\endgroup$
    – Undergrad2019
    Commented Oct 28, 2020 at 14:58
  • $\begingroup$ @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. $\endgroup$
    – Michael Seifert
    Commented Oct 28, 2020 at 17:06

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