From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ________ s.

Let height of tower be h and speed of projection in first two cases be u.

$$\begin{gathered} \text{F or case-I :}{2}^{\text{nd}} \text{equation}\mathrm{s}=\mathrm{ut}+\frac{1}{2} {\text{at}}^2 \\ \begin{array}{rl} & \mathrm{H}=-\mathrm{u}\left(6\right)+\frac{1}{2}\mathrm{g}(6{)}^2\\ & \mathrm{H}=-6\mathrm{u}+18\mathrm{g}\dots \left(\mathrm{i}\right)\end{array} \end{gathered}$$

$$\begin{gathered} \text{For case-II }:\mathrm{h}=\mathrm{u}(1.5)+\frac{1}{2}\mathrm{g}(1.5{)}^2 \\ \mathrm{h}=1.5\mathrm{u}+\frac{2.25\mathrm{g}}{2}\dots (\mathrm{ii}) \\ \mathrm{Multiplying} \mathrm{equation} (\mathrm{ii}) \mathrm{by} 4 \mathrm{we} \mathrm{get} \\ 4 \mathrm{h}=6\mathrm{u}+4.5 \mathrm{g}\dots .(\mathrm{iii}) \\ \mathrm{equation} (\mathrm{i}) + \mathrm{equation} (\mathrm{iii}) \mathrm{we} \mathrm{get} 5\mathrm{h} = 22.5\mathrm{g} \\ \mathrm{h}=4.5 \mathrm{g}\dots \text{ (iv) } \end{gathered}$$

$$\begin{gathered} \mathrm{For} \mathrm{case}-\mathrm{III} : \\ \mathrm{h}=0+\frac{{\displaystyle 1}}{{\displaystyle 2}}{\mathrm{gt}}^2\dots \left(\mathrm{v}\right) \\ \mathrm{Using} \mathrm{equation} \left(4\right) \& \mathrm{equation} \left(5\right) \\ \begin{array}{rl} & 4.5\mathrm{g}=\frac{1}{2}{\mathrm{gt}}^2\\ & {\mathrm{t}}^2=9\Rightarrow \mathrm{t}=3\mathrm{s}\end{array} \end{gathered}$$