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Optimizing Cannon Range on a Vertical Tower
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[QUOTE="RPinPA, post: 6071547, member: 651116"] And then what did you do. It looks like you applied the quadratic formula to the equation for ##y(t)## to solve for [I]t[/I] in terms of [I]y[/I]. Is that so? Why not say so? Why not explain your calculations instead of leaving us to guess if you want us to follow and comment on what you're doing? The remainder of your calculations seem to be following appropriate steps though I haven't tracked it in detail. Two comments on the final results: 1. ##\csc(\alpha) = 1/\sin(\alpha),## so if you have ##\sin(\alpha)## it's pretty easy to get ##\csc(\alpha)##. 2. You say the maximum range is when ##\sin(\alpha) = 1##, but that's when ##\alpha## is straight up and thus the range is 0. The cannonball lands on your head. Any angle less than that is going to give more range. So clearly there's an error somewhere. 3. There is something else wrong with your final steps. As I just said, if ##\sin(\alpha) = 1##, then the cannon is pointing directly upward. Also ##\cos(\alpha) = 0## which means x(t) = 0 for all t, so I don't know what steps you followed to get a nonzero ##x_{max}## from that. But clearly you're doing something invalid. I think I see the first place that seems to be going wrong anyway. That's an appropriate thing to do, though x may also have a minimum when ##dx/d\alpha = 0##. So if you find more than one such point, you'll need to check which is which. Why are you looking for where ##dx/d\alpha## is maximized? You just said you wanted to know where ##dx/d\alpha## is 0. [/QUOTE]
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Optimizing Cannon Range on a Vertical Tower
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