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A stone projected with a velocity a at an angle $\theta$ with the horizontal reaches maximum height $\mathrm{H}_{1}$. When it is projected with velocity u at an angle $\left(\frac{\pi}{2}-\theta\right)$ with the horizontal, it reaches maximum height $\mathrm{H}_{2}$. The relation between the horizontal range $\mathrm{R}$ of the projectile, heights $\mathrm{H}_{1}$ and $\mathrm{H}_{2}$ is
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$\mathrm{R}=4 \sqrt{\mathrm{H}_{1} \mathrm{H}_{2}}$
$\begin{aligned} & \mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}} \\ \text { and } \mathrm{H}_{2}=\frac{\mathrm{u}^{2} \sin ^{2}\left(90^{\circ}-\theta\right)}{2 \mathrm{~g}}=\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{~g}} \\ \mathrm{H}_{1} \mathrm{H}_{2}=& \frac{u^{2} \sin ^{2} \theta}{2 \mathrm{~g}} \times \frac{u^{2} \cos ^{2} \theta}{2 \mathrm{~g}}=\frac{\left(u^{2} \sin 2 \theta\right)^{2}}{16 g^{2}}=\frac{R^{2}}{16} \\ \therefore R=4 \sqrt{H_{1} H_{2}} \end{aligned}$
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