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 \(=0\)
\(v^{\prime 2}-v^2=2(a) \times(-1)\)
\(\Rightarrow -18 g=-2 a\)
\(\Rightarrow a=9 g\)