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

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${t_1}{t_2} \propto R$

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(B) For a given velocity $u$,the range $R$ is the same for two 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 the two times of flight:$t_1 t_2 = \left( \frac{2u \sin 	heta}{g} ight) \left( \frac{2u \cos 	heta}{g} ight) = \frac{4u^2 \sin 	heta \cos 	heta}{g^2}$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get:$t_1 t_2 = \frac{2u^2 (2 \sin 	heta \cos 	heta)}{g^2} = \frac{2(u^2 \sin 2	heta)}{g^2}$.Since the range $R = \frac{u^2 \sin 2	heta}{g}$,we can substitute this into the equation:$t_1 t_2 = \frac{2R}{g}$.Since $g$ is constant,we conclude that $t_1 t_2 \propto R$.

For a given velocity,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:

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