# Is it possible to guess this solution just by looking at the system? [closed]

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.

• 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? – Aaron Stevens Aug 9 '19 at 19:23
• @AaronStevens Yes – Brain Stroke Patient Aug 9 '19 at 19:25
• Are you familiar with the h =0 limit of the formula, and do you have an inspection argument for it? – Cosmas Zachos Aug 9 '19 at 19:49
• @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. – Brain Stroke Patient Aug 9 '19 at 19:52
• One of the answers here summarizes the geometrical construction best. – Cosmas Zachos Aug 9 '19 at 21:21