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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........ $s$.
$38$
$7$
$8$
$3$
Solution
Let height of tower be $h$ and speed of projection in first two cases be $u$.
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 $(5)$
$4.5 g=\frac{1}{2} g t^{2}$
$t ^{2}=9 \Rightarrow t =3 s$
