Answer
The stone crosses each other at a distance $2.46$ $m$ from the base of the cliff.
Work Step by Step
Let downward be the positive direction and upward be the negative direction.
Say, the stone crosses each other at a distance $x$ from the base of the cliff.
The equation of motion : $s=ut+\frac{1}{2}gt^{2}$
For upward motion:
$x=9t-\frac{1}{2}gt^{2} \tag{1}$
For downward motion:
$(6-x)=9t+\frac{1}{2}gt^{2} \tag{2}$
$x=6-9t-\frac{1}{2}gt^{2} \tag{3}$
$9t-\frac{1}{2}gt^{2}=6-9t-\frac{1}{2}gt^{2}$
$18t=6$
Equation equation $(1)$ and $(3)$,
$t=\frac{1}{3}$ $s$
Putting $t=\frac{1}{3}$ $s$ in equation $(1)$,
$x=9\times\frac{1}{3}-\frac{1}{2}\times9.8\times\frac{1}{3^{2}}$
$x=3-0.54$
$x=2.46$ $m$
