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(C) Let the initial velocities of the two stones be $u_1$ and $u_2$ $(u_2 > u_1)$.The positions of the stones at time $t$ are given by $x_1(t) = u_1 t

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(C) Let the initial velocities of the two stones be $u_1$ and $u_2$ $(u_2 > u_1)$.The positions of the stones at time $t$ are given by $x_1(t) = u_1 t - \frac{1}{2} g t^2$ and $x_2(t) = u_2 t - \frac{1}{2} g t^2$.The relative position is $\Delta x = |x_2(t) - x_1(t)| = |(u_2 - u_1)t| = (u_2 - u_1)t$.This shows that while both stones are in the air,the relative position $\Delta x$ increases linearly with time $t$.Once the first stone (with speed $u_1$) hits the ground at $t_1 = \frac{2u_1}{g}$,its position becomes $x_1 = 0$. The relative position becomes $\Delta x = |x_2(t) - 0| = |u_2 t - \frac{1}{2} g t^2|$.This is a downward-opening parabola. Thus,the graph should show a linear increase followed by a parabolic decrease until the second stone hits the ground. Graph $C$ correctly represents this behavior.

Two stones are thrown up vertically and simultaneously but with different speeds. Which graph correctly represents the time variation of their relative position $\Delta x$? Assume that stones do not bounce after hitting the ground.

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