1. Take upward as positive and origin at the top of the cliff

Let stone 1 be thrown upward with speed $u_1=10\,\text{m/s}$ and stone 2 with speed $u_2=40\,\text{m/s}$.

Their positions (while in air) are $y_1=10t-5t^2,$ $y_2=40t-5t^2.$

So the relative position of stone 2 with respect to stone 1 is $y_{21}=y_2-y_1=(40t-5t^2)-(10t-5t^2)=30t.$

Thus, as long as both stones are in air, the relative position increases linearly with time.

2. Find when each stone hits the ground

The ground is $240\,\text{m}$ below the launch point, so ground level is at $y=-240$.

For stone 1:

$10t-5t^2=-240$ $5t^2-10t-240=0$ $t^2-2t-48=0$ $(t-8)(t+6)=0$ So, $t_1=8\,\text{s}.$

For stone 2:

$40t-5t^2=-240$ $5t^2-40t-240=0$ $t^2-8t-48=0$ $t=12\,\text{s}.$

So stone 1 hits first at $8\,\text{s}$ and stone 2 hits later at $12\,\text{s}$.

3. Relative position in different time intervals

Case I: $0\le t\le 8$ s

Both are moving freely, so $y_{21}=30t.$ This is a straight line through origin with positive slope.

At $t=8$ s, $y_{21}=30\times 8=240\,\text{m}.$

Case II: $8\le t\le 12$ s

Stone 1 has already hit the ground and stays there at $y_1=-240$. Stone 2 is still moving, so $y_{21}=y_2-(-240)=40t-5t^2+240.$

This is a downward opening parabola.

Check values:

• At $t=8$: $y_{21}=40(8)-5(8)^2+240=320-320+240=240,$ so it joins smoothly.
• At $t=12$: $y_{21}=40(12)-5(12)^2+240=480-720+240=0.$

Thus after $t=8$ s, the relative position decreases parabolically from $240$ to $0$ at $t=12$ s.

4. Shape of the graph

So the graph must have:

1. A straight line increasing from $t=0$ to $t=8$ s.
2. Then a curved (concave downward) decrease from $t=8$ to $t=12$ s.
3. After $t=12$ s, both stones are on the ground, so relative position remains $0$.

This corresponds to Option A.

5. Comparison with stored answer

Derived answer: A

Stored correct answer: A

They agree.