The horizontal range is four times the maximu | 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

$45$

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 $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Given that the horizontal range is four times the maximum height,we have $R = 4H$.Substituting the formulas: $\frac{2u^2 \sin 	heta \cos 	heta}{g} = 4 \left( \frac{u^2 \sin^2 	heta}{2g} ight)$.Simplifying the equation: $\frac{2u^2 \sin 	heta \cos 	heta}{g} = \frac{2u^2 \sin^2 	heta}{g}$.This leads to $\cos 	heta = \sin 	heta$,which implies $	an 	heta = 1$.Therefore,$	heta = 45^\circ$.

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is ......... $^o$

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