A stone thrown with velocity ' $u$ ' at angles ' $ heta^{\prime}$ and $\left(90^{\circ}- heta ight)$ with the horizontal reaches to maximum heights $\mathrm{H}_1$ and $\mathrm{H}_2$ respectively. Its horizontal range is

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A stone thrown with velocity ' $u$ ' at angles ' $	heta^{\prime}$ and $\left(90^{\circ}-	hetaight)$ with the horizontal reaches to maximum heights $\mathrm{H}_1$ and $\mathrm{H}_2$ respectively. Its horizontal range is

Aicharya · Accepted Answer

Maximum height, $H=\frac{u^2 \sin ^2 	heta}{2 g}$ and horizontal range, $ R=\frac{u^2 \sin 2 	heta}{g} $ $\begin{aligned} & \mathrm{H}_1=\frac{\mathrm{u}^2 \sin ^2 	heta}{2 \mathrm{~g}} 	ext { and } \mathrm{H}_2=\frac{\mathrm{u}^2 \sin ^2\left(90^{\circ}-	hetaight)}{2 \mathrm{~g}}=\frac{\mathrm{u}^2 \cos ^2 	heta}{2 \mathrm{~g}} \ & 	herefore \mathrm{H}_1 \mathrm{H}_2=\frac{\mathrm{u}^2 \sin ^2 	heta}{2 \mathrm{~g}} 	imes \frac{\mathrm{u}^2 \cos ^2 	heta}{2 \mathrm{~g}}=\frac{\left(\mathrm{u}^2 \sin 2 	hetaight)^2}{16 \mathrm{~g}^2}=\frac{\mathrm{R}^2}{16} \ & 	herefore \mathrm{R}=4 \sqrt{\mathrm{H}_1 \mathrm{H}_2}\end{aligned}$

A stone thrown with velocity ' $u$ ' at angles ' $\theta^{\prime}$ and $\left(90^{\circ}-\theta\right)$ with…

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