A ball is thrown at an angle theta with the h | vedclass

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(A) 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

Vedclass · Accepted Answer

$4$

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(A) 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 $H$ of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.According to the problem,the horizontal range is equal to the maximum height,so $R = H$.Substituting the formulas: $\frac{2u^2 \sin 	heta \cos 	heta}{g} = \frac{u^2 \sin^2 	heta}{2g}$.Canceling common terms $(u^2/g)$: $2 \sin 	heta \cos 	heta = \frac{\sin^2 	heta}{2}$.Rearranging the terms: $4 \sin 	heta \cos 	heta = \sin^2 	heta$.Dividing both sides by $\sin 	heta \cos 	heta$ (assuming $	heta 
eq 0$): $4 = \frac{\sin 	heta}{\cos 	heta}$.Therefore,$	an 	heta = 4$.

$A$ ball is thrown at an angle $\theta$ with the horizontal. Its horizontal range is equal to its maximum height. This is possible only when the value of $\tan \theta$ is ..........

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