The ball's position follows the free fall equation:
\[
h = \frac{1}{2} g T^2
\]
Here, \( g \) is gravitational acceleration and \( T \) is the fall time.
We want the ball's height after \( T/5 \) seconds. The distance covered in \( T/5 \) seconds is:
\[
h' = \frac{1}{2} g \left( \frac{T}{5} \right)^2 = \frac{1}{2} g \frac{T^2}{25}
\]
So, the height above the ground at \( T/5 \) seconds is:
\[
\text{Height above the ground} = h - \frac{h}{25} = \frac{24h}{25}
\]
The answer is \( \frac{24h}{25} \, \text{m} \).