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(C) The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the initial velocity,$	heta$ is the a

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$R_1 = R_2$

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(C) The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the initial velocity,$	heta$ is the angle of projection with the horizontal,and $g$ is the acceleration due to gravity.For the first case,the angle with the horizontal is $	heta$,so the range is $R_1 = \frac{u^2 \sin(2	heta)}{g}$.For the second case,the angle with the vertical is $	heta$,which means the angle with the horizontal is $(90^{\circ} - 	heta)$. Therefore,the range is $R_2 = \frac{u^2 \sin(2(90^{\circ} - 	heta))}{g} = \frac{u^2 \sin(180^{\circ} - 2	heta)}{g}$.Since $\sin(180^{\circ} - 2	heta) = \sin(2	heta)$,we have $R_2 = \frac{u^2 \sin(2	heta)}{g}$.Comparing the two expressions,we find that $R_1 = R_2$.

$A$ projectile is thrown at an angle $\theta$ with the horizontal and its range is $R_1$. It is then thrown at an angle $\theta$ with the vertical and the range is $R_2$. Then:

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