h મીટર ઊંચાઇના ટાવર પરથી એક દડાને મુકત કરવ

$\;\;h = ut + \frac{1}{2}g{t^2} \Rightarrow h = \frac{1}{2}g{T^2}$

After $\frac{T}{3}$ seconds, the position of ball,

$h' = 0 + \frac{1}{2}g{\left( {\frac{T}{3}} \right)^2} = \frac{1}{2} \times \frac{g}{9} \times {T^2}$

$h' = \frac{1}{2} \times \frac{g}{9} \times {T^2}$$ = \frac{h}{9}\,m$ from top

Position of ball from ground $ = h - \frac{h}{9} = \frac{{8\;h}}{9}\,m.$