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

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Four times the vertical height

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(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.The maximum vertical height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Dividing $R$ by $H$,we get $\frac{R}{H} = \frac{u^2 \sin(2	heta) / g}{u^2 \sin^2 	heta / (2g)} = \frac{2 \sin 	heta \cos 	heta}{\sin^2 	heta / 2} = 4 \cot 	heta$.Therefore,$R = 4H \cot 	heta$.Given the angle of projection $	heta = 45^\circ$,we substitute this into the equation:$R = 4H \cot(45^\circ)$.Since $\cot(45^\circ) = 1$,we have $R = 4H$.Thus,the horizontal range is four times the vertical height.

An object is thrown along a direction inclined at an angle of $45^\circ$ with the horizontal direction. The horizontal range of the particle is equal to

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