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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 $\left(\frac{\pi}{2}-\theta\right)$ 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
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The correct answer is:
$R=4 \sqrt{H_1 H_2}$
We know that, $H_1=\frac{u^2 \sin ^2 \theta}{2 g}$
$\begin{aligned} & \text { and } \quad H_2=\frac{u^2 \sin ^2(90-\theta)}{2 g} \\ & \Rightarrow \quad H_2=\frac{u^2 \cos ^2 \theta}{2 g} \\ & \because \quad H_1 H_2=\frac{u^2 \sin ^2 \theta}{2 g} \times \frac{u^2 \cos ^2 \theta}{2 g}\end{aligned}$
$=\frac{4}{4} \times \frac{U^2 \sin ^2 \theta}{2 g} \times \frac{U^2 \cos ^2 \theta}{2 g}$
$=\frac{\left(u^2 \sin 2 \theta\right)^2}{16 g^2} \quad\left[\because \sin ^2 \theta=2 \sin \theta \cos \theta\right]$
$\begin{aligned} & =\frac{R^2}{16} \\ R^2 & =6 H_1 H_2 \\ \therefore \quad R & =4 \sqrt{H_1 H_2}\end{aligned}$
$\begin{aligned} & \text { and } \quad H_2=\frac{u^2 \sin ^2(90-\theta)}{2 g} \\ & \Rightarrow \quad H_2=\frac{u^2 \cos ^2 \theta}{2 g} \\ & \because \quad H_1 H_2=\frac{u^2 \sin ^2 \theta}{2 g} \times \frac{u^2 \cos ^2 \theta}{2 g}\end{aligned}$
$=\frac{4}{4} \times \frac{U^2 \sin ^2 \theta}{2 g} \times \frac{U^2 \cos ^2 \theta}{2 g}$
$=\frac{\left(u^2 \sin 2 \theta\right)^2}{16 g^2} \quad\left[\because \sin ^2 \theta=2 \sin \theta \cos \theta\right]$
$\begin{aligned} & =\frac{R^2}{16} \\ R^2 & =6 H_1 H_2 \\ \therefore \quad R & =4 \sqrt{H_1 H_2}\end{aligned}$
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