Question

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 ___$S$.

Answer 1

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) \& equation (5)
$4.5 g =\frac{1}{2} gt ^{2}$
$t^{2}=9 $
$\Rightarrow t=3\, s$