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(B) For a projectile,the range $R$ is the same for two complementary angles of projection,$	heta$ and $(90^\circ - 	heta)$.The time of flight for an

Vedclass · Accepted Answer

$\frac{T_1T_2g}{2}$

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(B) For a projectile,the range $R$ is the same for two complementary angles of projection,$	heta$ and $(90^\circ - 	heta)$.The time of flight for angle $	heta$ is $T_1 = \frac{2u \sin 	heta}{g}$.The time of flight for angle $(90^\circ - 	heta)$ is $T_2 = \frac{2u \sin(90^\circ - 	heta)}{g} = \frac{2u \cos 	heta}{g}$.Multiplying $T_1$ and $T_2$,we get:$T_1 T_2 = \left(\frac{2u \sin 	heta}{g}ight) \left(\frac{2u \cos 	heta}{g}ight) = \frac{4u^2 \sin 	heta \cos 	heta}{g^2} = \frac{2u^2 (2 \sin 	heta \cos 	heta)}{g^2} = \frac{2u^2 \sin 2	heta}{g^2}$.Since the range $R = \frac{u^2 \sin 2	heta}{g}$,we can substitute this into the equation:$T_1 T_2 = \frac{2R}{g}$.Therefore,$R = \frac{T_1 T_2 g}{2}$.

$A$ projectile has the same range $R$ for two angles of projection. If $T_1$ and $T_2$ are the times of flight in the two cases,then $R$ is

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