\[(a)\;\frac{u^2\cos \alpha}{2g} \quad (b)\;\frac{u^2\sin \alpha}{2g}\quad (c)\;\frac{u^2}{g} \quad (d)\;\frac{u^2}{2g}\]

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The two angles of projection which the horizontal range is same are $\alpha$ and $90-\alpha$

$H_1=\large\frac{u^2 \sin ^ 2 \alpha}{2g}$

$H_2=\large\frac{u^2 \sin ^ 2 (90-\alpha)}{2g}$

$\qquad=\large\frac{u^2 \cos^2 \alpha}{2g}$

$H_1+H_2=\large\frac{u^2}{2g}$$(\sin ^2 \alpha+\cos ^2 \alpha)$

$\qquad=\large\frac{u^2}{2g}$

Hence d is the correct answer.

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