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(D) The time of flight for a projectile is given by $T = \frac{2v \sin 	heta}{g}$. Since $v_1 \sin 	heta_1 = v_2 \sin 	heta_2$,it follows that $T_1

Vedclass · Accepted Answer

None of these.

Vedclass · Answer

(D) The time of flight for a projectile is given by $T = \frac{2v \sin 	heta}{g}$. Since $v_1 \sin 	heta_1 = v_2 \sin 	heta_2$,it follows that $T_1 = T_2$. Thus,option $A$ is correct.The maximum height is given by $H = \frac{v^2 \sin^2 	heta}{2g}$. Since $v_1 \sin 	heta_1 = v_2 \sin 	heta_2$,it follows that $H_1 = H_2$. Thus,option $B$ is correct.The relative acceleration between the two projectiles is $\vec{a}_{rel} = \vec{g} - \vec{g} = 0$. Since the relative acceleration is zero,the relative velocity $\vec{v}_{rel}$ remains constant. The relative vertical velocity is $v_{1y} - v_{2y} = v_1 \sin 	heta_1 - v_2 \sin 	heta_2 = 0$. Since the relative vertical velocity is zero,the relative motion is purely horizontal. Thus,the trajectory of one with respect to the other is a horizontal straight line. Option $C$ is correct.Since all statements $A$,$B$,and $C$ are correct,the incorrect statement is none of these.

Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are $v_1$ and $v_2$ at angles $\theta_1$ and $\theta_2$ respectively from the horizontal,and $v_1 \sin \theta_1 = v_2 \sin \theta_2$,then choose the incorrect statement.

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