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(C) Let the initial velocity of body $A$ be $u$. Since it reaches a maximum height $h$,we have $u^2 = 2gh$,so $u = \sqrt{2gh}$.For body $A$,the veloci

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(C) Let the initial velocity of body $A$ be $u$. Since it reaches a maximum height $h$,we have $u^2 = 2gh$,so $u = \sqrt{2gh}$.For body $A$,the velocity at time $t$ is $V_A = u - gt$.For body $B$,which is dropped from height $h$,the velocity at time $t$ is $V_B = -gt$ (taking downward as negative).Relative velocity $V_{AB} = V_A - V_B = (u - gt) - (-gt) = u$.This holds until body $B$ strikes the ground. The time taken for $B$ to fall height $h$ is $t_0 = \sqrt{2h/g}$.Since $u = \sqrt{2gh}$,we have $t_0 = \sqrt{2h/g} = \sqrt{2h/g} = u/g$.So for $0 \le t \le t_0$,$V_{AB} = u$ (a constant positive value).After $t > t_0$,body $B$ stops $(V_B = 0)$,so $V_{AB} = V_A = u - gt$.This is a linear equation with a negative slope $-g$.Thus,the graph shows a constant positive value until $t_0$,followed by a downward sloping line. This matches the representation in option $C$.

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