A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in $\frac{T}{3}$ seconds

The correct option is C 8h/9 meters from the ground

$\because$ 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

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