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For angle ...X..., the projectile has maximum range and it is equal to ...X.... Here, X and Y refer to.
Explanation
If the angle of projection is $\pi/4$, then range = $\frac{v_{0}^{2}}{g}\sin(\pi/2)$ $\Rightarrow (R)_{max} = \frac{v_{0}^{2}}{g} [ \because \{\sin(\pi/2)\}_{max} = 1]$