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

Vedclass · Accepted Answer

$	heta = 	an^{-1}(4)$

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(C) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.The maximum height $H$ of a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.According to the problem,$R = H$.Therefore,$\frac{u^2 \sin 2	heta}{g} = \frac{u^2 \sin^2 	heta}{2g}$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get:$\frac{2 u^2 \sin 	heta \cos 	heta}{g} = \frac{u^2 \sin^2 	heta}{2g}$.Canceling $u^2$,$g$,and $\sin 	heta$ (assuming $\sin 	heta 
eq 0$):$2 \cos 	heta = \frac{\sin 	heta}{2}$.Rearranging the terms,we get $\frac{\sin 	heta}{\cos 	heta} = 4$,which means $	an 	heta = 4$.Thus,$	heta = 	an^{-1}(4)$.

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

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