The circus girl catches the first ring after \\\\(\\dfrac{1}{2}+\\dfrac{1}{2}+\\dfrac{1}{2}=\\dfrac{3}{2}\\\\) second\\\\(\\Rightarrow \\\\) the ring was in air for\\\\(\\dfrac{3}{2} \\\\) second in which for first \\\\(\\dfrac{3}{4} \\\\) second it was going up and for the next \\\\(\\dfrac{3}{4} \\\\) second it was coming down.Let the height attained by first ring be h in \\\\(\\dfrac{3}{4} \\\\) second, obviously, its velocity at the instant will be zero.if u be the velocity of projection of rings, then\\\\(0=u-g\\left ( \\dfrac{3}{4} \\right )\\Rightarrow u=\\dfrac{3g}{4}\\\\)\\\\(\\Rightarrow (A)\\\\) is correct.The maximum height h attained by the ring is given by,\\\\(0=\\left ( \\dfrac{3g}{4} \\right )^{2}-2gh\\\\)\\\\(\\Rightarrow h=\\dfrac{9g}{32}\\Rightarrow (B)\\\\) is not correct.When the first ring returns to her hand, the second ring has traveled for 1 second. Of that 1 second, \\\\(\\dfrac{3}{4}\\\\) seconds will be spent in going up and for the rest it is coming down.\\\\(\\Rightarrow h'=\\dfrac{1}{2}g\\left ( \\dfrac{1}{4} \\right )^{2}=\\dfrac{g}{32}\\\\)\\\\(\\Rightarrow \\\\) Distance travelled by second ring by that instant is\\\\(=\\dfrac{9g}{32}-\\dfrac{g}{32}=\\dfrac{g}{4}\\\\)\\\\(\\Rightarrow \\\\) (C) is correct.The third ring has travelled for \\\\(t=\\dfrac{1}{2}\\\\) second by that instant. it will be going up.Corresponding distance from ground\\\\(t=\\left ( \\dfrac{3g}{4} \\right )\\left ( \\dfrac{1}{2} \\right )-\\dfrac{1}{2}g\\left ( \\dfrac{1}{2} \\right )^{2}\\\\)\\\\(=\\dfrac{3g}{8}-\\dfrac{g}{8}=\\dfrac{g}{4}\\\\)\\\\(\\Rightarrow (D)\\\\) is correct.