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(B) The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$.Since the initial speed $u$ is the same for all thr

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$R_1 = R_3 < R_2$

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(B) The horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$.Since the initial speed $u$ is the same for all three balls,the range depends on $\sin(2	heta)$.For $	heta_1 = 30^{\circ}$,$R_1 \propto \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866$.For $	heta_2 = 45^{\circ}$,$R_2 \propto \sin(90^{\circ}) = 1$.For $	heta_3 = 60^{\circ}$,$R_3 \propto \sin(120^{\circ}) = \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866$.Since $30^{\circ} + 60^{\circ} = 90^{\circ}$,the ranges for complementary angles are equal,so $R_1 = R_3$.Since $\sin(90^{\circ})$ is the maximum value of the sine function,$R_2$ is the maximum range.Therefore,$R_1 = R_3 < R_2$.

Three identical balls are projected with the same speed at angles $30^{\circ}$,$45^{\circ}$,and $60^{\circ}$. Their ranges are $R_1$,$R_2$,and $R_3$ respectively. Then:

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