Question

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

Answer 1

Answer: 3

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

For case-I : $2^{\text{nd }}$ equation $s=ut+\frac{1}{2}a{t}^2$
$h=-u(6)+\frac{1}{2}g(6{)}^2$
$H=-6u+18g\dots$ (i)
For case-II : $h=u(1.5)+\frac{1}{2}g(1.5{)}^2$
$h=1.5u+\frac{2.25g}{2}\dots$ (ii)
Multiplying equation (ii) by $4$ we get
$4h=6u+4.5g\dots$. (iii)
equation (i) $+$ equation (iii) we get
$5h=22.5g$ $h=4.5g\dots$ (iv)
For case-III :
$h=0+\frac{1}{2}g{t}^2\dots (v)$
Using equation (4) \& equation (5)
$4.5g=\frac{1}{2}g{t}^2$
$t^2=9$
$\Rightarrow t=3s$