A projectile can have the same range R for tw | vedclass

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(C) For a projectile with initial velocity $u$ and angle of projection $	heta$,the range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$.For the sa

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$R$

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(C) For a projectile with initial velocity $u$ and angle of projection $	heta$,the range $R$ is given by $R = \frac{u^2 \sin(2	heta)}{g}$.For the same range $R$,the two angles of projection are $	heta$ and $(90^\circ - 	heta)$.The time of flight $t$ is given by $t = \frac{2u \sin	heta}{g}$.For the first angle $	heta$,$t_1 = \frac{2u \sin	heta}{g}$.For the second angle $(90^\circ - 	heta)$,$t_2 = \frac{2u \sin(90^\circ - 	heta)}{g} = \frac{2u \cos	heta}{g}$.The product of the times of flight is $t_1 t_2 = \left(\frac{2u \sin	heta}{g}ight) \left(\frac{2u \cos	heta}{g}ight) = \frac{4u^2 \sin	heta \cos	heta}{g^2}$.Using the identity $\sin(2	heta) = 2 \sin	heta \cos	heta$,we get $t_1 t_2 = \frac{2u^2 \sin(2	heta)}{g^2}$.Since $R = \frac{u^2 \sin(2	heta)}{g}$,we substitute this into the equation to get $t_1 t_2 = \frac{2R}{g}$.Therefore,$t_1 t_2 \propto R$.

$A$ projectile can have 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 the product of the two times of flight is proportional to

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