a
\(\begin{array}{l}
h = \frac{1}{2}g{T^2}\\
now\,for\,t = T/3\,{\rm{second}}\,vertical\,{\rm{distance}}\,\\
moved\,is\,given\,by\\
h' = \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} \Rightarrow h' = \frac{1}{2} \times \frac{{g{T^2}}}{9} = \frac{h}{9}\\
\therefore \,Position\,of\,ball\,from\,ground\, = h - \frac{h}{9}\\
= \frac{{8h}}{9}
\end{array}\)