The angle of projection at which the horizont | vedclass

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(C) 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

$	heta = 	an^{-1}(4)$ or $	heta \approx 76^\circ$

Vedclass · Answer

(C) 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 $H = \frac{u^2 \sin^2	heta}{2g}$.Given that the horizontal range is equal to the maximum height,$R = H$.Substituting the formulas: $\frac{2u^2 \sin	heta \cos	heta}{g} = \frac{u^2 \sin^2	heta}{2g}$.Simplifying the equation: $2 \cos	heta = \frac{\sin	heta}{2}$.Rearranging for $	an	heta$: $\frac{\sin	heta}{\cos	heta} = 4$,which means $	an	heta = 4$.Therefore,$	heta = 	an^{-1}(4)$,which is approximately $76^\circ$.

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

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