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Show that, if a projectile is shot from a height $h$ with speed $v_0$, the maximum range is obtained for launch angle $\theta=\arctan\left(\dfrac{v_0}{\sqrt{2gh+v_0^2}}\right)$

For the problem shown above, I noticed how the numerator is the initial speed and the denominator is the final speed, and so I wonder if there's a way to solve this problem through some sort of physical insight and simple geometry maybe? I tried thinking for quite a while but I couldn't find an answer as to why the maximum range angle is the arctan of the ratio of the start and end speeds.

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  • $\begingroup$ To clarify, is this saying that you need to launch at this angle to get the maximum range. i.e. the farthest horizontal distance before hitting the ground? $\endgroup$ – Aaron Stevens Aug 9 '19 at 19:23
  • $\begingroup$ @AaronStevens Yes $\endgroup$ – Brain Stroke Patient Aug 9 '19 at 19:25
  • $\begingroup$ Are you familiar with the h =0 limit of the formula, and do you have an inspection argument for it? $\endgroup$ – Cosmas Zachos Aug 9 '19 at 19:49
  • $\begingroup$ @CosmasZachos I don't know what you mean by an inspection argument but yes, I did notice that when h = 0 , the maximum range angle comes out to be 45 degrees as expected, if that's what you meant. $\endgroup$ – Brain Stroke Patient Aug 9 '19 at 19:52
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    $\begingroup$ One of the answers here summarizes the geometrical construction best. $\endgroup$ – Cosmas Zachos Aug 9 '19 at 21:21