the acceleration of the ball will be g. Initial velocity will be 0. in T sec. body travels h mts. by applying equations of motion we get\\\\(s= ut +(1/2)gT^2\\\\)\\\\(h = (1/2)gT^2\\\\) ------[1]in \\\\(T/3\\\\) sec \\\\(h_1 = (1/2)gT^2=(1/2)g(\\dfrac{T}{3})^2=(1/2)g(\\dfrac{T^2}{9})\\\\) -------[2]from \\\\([1]\\\\) and \\\\([2]\\\\) we get \\\\(h_1 =\\dfrac{h}{9}\\\\) distance from point of release.therefore distance from ground is \\\\(h-\\dfrac{h}{9} =\\dfrac{8h}{9}\\\\)