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(A) The time of flight for a projectile is given by $T = \frac{2v \sin 	heta}{g}$.For two complementary angles $	heta$ and $(90^\circ - 	heta)$, th

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$250$

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(A) The time of flight for a projectile is given by $T = \frac{2v \sin 	heta}{g}$.For two complementary angles $	heta$ and $(90^\circ - 	heta)$, the horizontal range $R$ is the same.The times of flight are $T_1 = \frac{2v \sin 	heta}{g} = 5 	ext{ s}$ and $T_2 = \frac{2v \sin(90^\circ - 	heta)}{g} = \frac{2v \cos 	heta}{g} = 10 	ext{ s}$.The horizontal range is $R = \frac{v^2 \sin(2	heta)}{g} = \frac{2v^2 \sin 	heta \cos 	heta}{g}$.Multiplying $T_1$ and $T_2$, we get $T_1 T_2 = \frac{4v^2 \sin 	heta \cos 	heta}{g^2} = \frac{2}{g} \left( \frac{2v^2 \sin 	heta \cos 	heta}{g} ight) = \frac{2R}{g}$.Therefore, $R = \frac{1}{2} g T_1 T_2 = \frac{1}{2} 	imes 10 	imes 5 	imes 10 = 250 	ext{ m}$.

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$. If the time of flight of these bodies are $5 \text{ s}$ and $10 \text{ s}$, respectively, then the value of $R$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

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