A stone projected with a velocity $u$ at an angle $\theta$ with the horizontal reaches maximum height $H_{1}$. When it is projected with velocity $u$ at an angle $\left(\frac{\pi}{2} - 0\right)$ with the horizontal, it reaches maximum height $H_{2}$. The relation between the horizontal range $R$ of the projectile, $H_{1}$ and $H_{2}$ is

For projection at angle $\theta$, the maximum height is $$H_1=\frac{u^2\sin^2\theta}{2g}.$$ For projection at angle $\left(\frac{\pi}{2}-\theta\right)$, the maximum height is $$H_2=\frac{u^2\sin^2\left(\frac{\pi}{2}-\theta\right)}{2g}=\frac{u^2\cos^2\theta}{2g}.$$ Multiplying, $$H_1H_2=\frac{u^4\sin^2\theta\cos^2\theta}{4g^2}.$$ Hence $$4\sqrt{H_1H_2}=4\cdot \frac{u^2\sin\theta\cos\theta}{2g}=\frac{2u^2\sin\theta\cos\theta}{g}=\frac{u^2\sin 2\theta}{g}.$$ But the horizontal range is $$R=\frac{u^2\sin 2\theta}{g}.$$ Therefore $$R=4\sqrt{H_1H_2}.$$ This matches option $\text{A}$.