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(A) Let the positions of the two stones be $x_1(t)$ and $x_2(t)$.For $0 \le t \le 6 \, s$,both stones are in the air under gravity $g$. Their position

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(A) Let the positions of the two stones be $x_1(t)$ and $x_2(t)$.For $0 \le t \le 6 \, s$,both stones are in the air under gravity $g$. Their positions are $x_1(t) = u_1 t - \frac{1}{2} g t^2$ and $x_2(t) = u_2 t - \frac{1}{2} g t^2$.The relative position is $\Delta x = x_2 - x_1 = (u_2 - u_1) t$. Since $u_2 > u_1$,this is a linear function of $t$ starting from $0$ at $t = 0$ and increasing until $t = 6 \, s$.At $t = 6 \, s$,the first stone hits the ground,so $x_1(6) = 0$. For $6 \, s The relative position is $\Delta x = x_2 - 0 = u_2 t - \frac{1}{2} g t^2$. This is a downward-opening parabola.At $t = 10 \, s$,the second stone hits the ground,so $x_2(10) = 0$,making $\Delta x = 0$.Thus,the graph is a straight line from $t = 0$ to $t = 6 \, s$ and a parabolic curve from $t = 6 \, s$ to $t = 10 \, s$. This matches the first graph.

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