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(C) Let the cliff edge be the origin $(y = 0)$ and the upward direction be positive. The positions of the two stones at time $t$ are given by:$y_1 = 1

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(C) Let the cliff edge be the origin $(y = 0)$ and the upward direction be positive. The positions of the two stones at time $t$ are given by:$y_1 = 10t - 5t^2$$y_2 = 40t - 5t^2$For the first stone,it hits the ground when $y_1 = -240 \ m$:$-240 = 10t - 5t^2 \implies t^2 - 2t - 48 = 0 \implies (t-8)(t+6) = 0$. Thus,$t = 8 \ s$.For $t \le 8 \ s$,the relative position is $y_{rel} = y_2 - y_1 = (40t - 5t^2) - (10t - 5t^2) = 30t$. This is a linear graph passing through the origin.At $t = 8 \ s$,$y_{rel} = 30(8) = 240 \ m$.For $t > 8 \ s$,the first stone is at rest on the ground $(y_1 = -240 \ m)$. The second stone hits the ground when $y_2 = -240 \ m$:$-240 = 40t - 5t^2 \implies t^2 - 8t - 48 = 0 \implies (t-12)(t+4) = 0$. Thus,$t = 12 \ s$.For $8 \ s Therefore,the graph is linear for $t \le 8 \ s$ and parabolic for $8 \ s < t \le 12 \ s$.

Column $I$	Column $II$
$(A)$ Change in velocity	$(p)$ $-5/3\,SI \text{ unit}$
$(B)$ Average acceleration	$(q)$ $-20\,SI \text{ unit}$
$(C)$ Total displacement	$(r)$ $-10\,SI \text{ unit}$
$(D)$ Acceleration at $t=3\,s$	$(s)$ $-5\,SI \text{ unit}$

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