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(D) The range of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.Since the ranges are equal,$\sin(2	heta_1) = \sin(2	heta_2)$.This implie

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$	an 	heta_1 / 	an 	heta_2$

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(D) The range of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g}$.Since the ranges are equal,$\sin(2	heta_1) = \sin(2	heta_2)$.This implies $2	heta_1 = 180^\circ - 2	heta_2$,which simplifies to $	heta_1 + 	heta_2 = 90^\circ$,or $	heta_2 = 90^\circ - 	heta_1$.The time of flight is given by $T = \frac{2u \sin 	heta}{g}$.The ratio of the times of flight is $\frac{T_1}{T_2} = \frac{\frac{2u \sin 	heta_1}{g}}{\frac{2u \sin 	heta_2}{g}} = \frac{\sin 	heta_1}{\sin 	heta_2}$.Since $	heta_2 = 90^\circ - 	heta_1$,we have $\sin 	heta_2 = \sin(90^\circ - 	heta_1) = \cos 	heta_1$.Thus,$\frac{T_1}{T_2} = \frac{\sin 	heta_1}{\cos 	heta_1} = 	an 	heta_1$.

If two bodies $A$ and $B$ are projected with the same velocity $u$ but with different angles $\theta_1$ and $\theta_2$ respectively with the horizontal such that both have the same range,then the ratio of the times of flight of the bodies $A$ and $B$ is:

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