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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

Vedclass · Accepted Answer

$53^\circ 8'$

Vedclass · Answer

(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 $H = \frac{u^2 \sin^2 	heta}{2g}$.According to the problem,the horizontal range is three times the maximum height,so $R = 3H$.Substituting the formulas:$\frac{2u^2 \sin 	heta \cos 	heta}{g} = 3 \left( \frac{u^2 \sin^2 	heta}{2g} ight)$.Simplifying the equation:$2 \cos 	heta = \frac{3}{2} \sin 	heta$.Rearranging to find $	an 	heta$:$	an 	heta = \frac{4}{3}$.Therefore,$	heta = 	an^{-1}(1.333) \approx 53^\circ 8'$.

$A$ body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is

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