If the ball is dropped then $x=0$, the velocity with which it will hit the sand will be given by

$v^2-u^2=2(-g)(-9)$

$v^2-0=18 g$

$v^2=18 g$

Now on striking sand, the body penetrates into sand for $1 m$ and comes to rest. So, $v \rightarrow$ initial for sand and final velocity $=0$

$v^{\prime 2}-v^2=2(a) \times(-1)$

$\Rightarrow -18 g=-2 a$

$\Rightarrow a=9 g$