The angle of projection for a projectile to h | vedclass

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(B) The formula for horizontal range $R$ is $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The formula for maximum height $

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$	an ^{-1}(4)$

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(B) The formula for horizontal range $R$ is $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The formula for maximum height $H$ is $H = \frac{u^2 \sin^2 	heta}{2g}$.Given that the range is equal to the maximum height,we set $R = H$:$\frac{2u^2 \sin 	heta \cos 	heta}{g} = \frac{u^2 \sin^2 	heta}{2g}$.Canceling common terms $u^2/g$ from both sides:$2 \sin 	heta \cos 	heta = \frac{\sin^2 	heta}{2}$.Dividing both sides by $\sin 	heta$ (assuming $\sin 	heta 
eq 0$):$2 \cos 	heta = \frac{\sin 	heta}{2}$.Rearranging to solve for $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$:$4 = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$.Therefore,$	heta = 	an^{-1}(4)$.

The angle of projection for a projectile to have the same horizontal range and maximum height is

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