A projectile can have the same range (R) for | vedclass

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(D) For the same range,$R_1 = R_2 = R$. Let the two angles of projection be $	heta$ and $(90^{\circ} - 	heta)$. The time of flight for a projectile

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$R$

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(D) For the same range,$R_1 = R_2 = R$. Let the two angles of projection be $	heta$ and $(90^{\circ} - 	heta)$. The time of flight for a projectile is given by $T = \frac{2u \sin 	heta}{g}$. For the first angle $	heta_1 = 	heta$,the time of flight is $T_1 = \frac{2u \sin 	heta}{g}$. For the second angle $	heta_2 = (90^{\circ} - 	heta)$,the time of flight is $T_2 = \frac{2u \sin(90^{\circ} - 	heta)}{g} = \frac{2u \cos 	heta}{g}$. The product of the times of flight is $T_1 T_2 = \left(\frac{2u \sin 	heta}{g}ight) \left(\frac{2u \cos 	heta}{g}ight)$. $T_1 T_2 = \frac{2}{g} \left(\frac{u^2 (2 \sin 	heta \cos 	heta)}{g}ight)$. Since the range $R = \frac{u^2 \sin 2	heta}{g}$,we have $T_1 T_2 = \frac{2R}{g}$. Since $g$ is constant,$T_1 T_2 \propto R$.

$A$ projectile can have the same range $(R)$ for two angles of projection. Their initial velocities are same. If $T_1$ and $T_2$ are times of flight in two cases,then the product of two times of flight is directly proportional to

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