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(C) The horizontal components of the velocities are given as $u_{x1} = v_1 \cos 	heta_1$ and $u_{x2} = v_2 \cos 	heta_2$. Given $v_1 \cos 	heta_1 =

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both will have the same range

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(C) The horizontal components of the velocities are given as $u_{x1} = v_1 \cos 	heta_1$ and $u_{x2} = v_2 \cos 	heta_2$. Given $v_1 \cos 	heta_1 = v_2 \cos 	heta_2$,the horizontal velocities are equal. Since the horizontal acceleration for both projectiles is zero,their horizontal positions at any time $t$ are $x_1 = (v_1 \cos 	heta_1)t$ and $x_2 = (v_2 \cos 	heta_2)t$. Because $v_1 \cos 	heta_1 = v_2 \cos 	heta_2$,it follows that $x_1 = x_2$ for all $t$. This means one particle will always remain exactly below or above the other particle (Option $A$ is correct). The relative horizontal displacement is $x_{rel} = x_1 - x_2 = 0$. Thus,the trajectory of one with respect to the other is a vertical straight line (Option $B$ is correct). The range of a projectile is $R = \frac{v^2 \sin(2	heta)}{g} = \frac{2(v \cos 	heta)(v \sin 	heta)}{g}$. Since $v_1 \cos 	heta_1 = v_2 \cos 	heta_2$,the ranges are equal only if $v_1 \sin 	heta_1 = v_2 \sin 	heta_2$,which is not necessarily true. Therefore,the statement that both will have the same range is not necessarily true. Thus,Option $C$ is the incorrect statement.

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