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(A) Let the height of the tower be $h$ and the speed of projection in the first two cases be $u$. Taking the downward direction as positive.For case-$

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$3$

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(A) Let the height of the tower be $h$ and the speed of projection in the first two cases be $u$. Taking the downward direction as positive.For case-$I$ (thrown upward): The displacement is $h$. The initial velocity is $-u$. The acceleration is $g$.Using the equation $s = ut + \frac{1}{2}at^2$:$h = -u(6) + \frac{1}{2}g(6)^2$$h = -6u + 18g$ ... $(i)$For case-$II$ (thrown downward): The displacement is $h$. The initial velocity is $u$. The acceleration is $g$.$h = u(1.5) + \frac{1}{2}g(1.5)^2$$h = 1.5u + 1.125g$ ... $(ii)$To eliminate $u$,multiply equation $(ii)$ by $4$:$4h = 6u + 4.5g$ ... $(iii)$Adding equation $(i)$ and equation $(iii)$:$h + 4h = (-6u + 18g) + (6u + 4.5g)$$5h = 22.5g$$h = 4.5g$ ... $(iv)$For case-$III$ (released from rest): The initial velocity is $0$. The acceleration is $g$.$h = 0(t) + \frac{1}{2}gt^2$$h = \frac{1}{2}gt^2$ ... $(v)$Substituting equation $(iv)$ into equation $(v)$:$4.5g = \frac{1}{2}gt^2$$t^2 = 9$$t = 3 \, s$.

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