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........ \(s\).

For case-I : \(2^{\text {nd }}\) equation \(s=u t+\frac{1}{2}\) at \(^{2}\)

\(h =- u (6)+\frac{1}{2}\, g (6)^{2}\)

\(H =-6 u +18\,g \ldots\) \((i)\)

For case-II \(: h = u (1.5)+\frac{1}{2}\, g (1.5)^{2}\)

\(h =1.5 u +\frac{2.25 g }{2} \ldots\) \((ii)\)

Multiplying equation \((ii)\) by \(4\) we get

\(4 h =6 u +4.5\, g \ldots .\) \((iii)\)

equation \((i)\) \(+\) equation \((iii)\) we get \(5 h =22.5 g\)

\(h =4.5 g \ldots\) \((iv)\)

For case\(-III\)

\(h =0+\frac{1}{2} gt ^{2} \ldots\) \((v)\)

Using equation \((4)\)and equation \((5)\)

\(4.5 g=\frac{1}{2} g t^{2}\)

\(t ^{2}=9 \Rightarrow t =3 s\)