The horizontal range of a projectile is 4sqrt | vedclass

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(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The maximum height

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$30$

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(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Dividing $R$ by $H$,we get $\frac{R}{H} = \frac{2u^2 \sin 	heta \cos 	heta}{g} 	imes \frac{2g}{u^2 \sin^2 	heta} = 4 \cot 	heta$.Given that $R = 4\sqrt{3} H$,we have $\frac{R}{H} = 4\sqrt{3}$.Equating the two expressions: $4 \cot 	heta = 4\sqrt{3}$.Therefore,$\cot 	heta = \sqrt{3}$.Since $\cot 30^\circ = \sqrt{3}$,the angle of projection $	heta = 30^\circ$.

The horizontal range of a projectile is $4\sqrt{3}$ times its maximum height. Its angle of projection will be ......... $^\circ$.

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