A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of ball above the ground in T/5 seconds?

Options

A. `(h)/(25)m`
B. 25 h m
C. 24 h m
D. `(24)/(25)` h m

Explanation

To solve the problem, we need to determine the position of the ball above the ground after \( \frac{T}{5} \) seconds when it is released from a height \( h \) meters and takes \( T \) seconds to reach the ground.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The ball is dropped from a height \( h \) and falls freely under the influence of gravity. The distance fallen by the ball after time \( t \) is given by the equation:
\[ x = \frac{1}{2} g t^2 \]
where \( g \) is the acceleration due to gravity.

2. **Total Time to Fall**:
The total time taken to reach the ground is \( T \). Therefore, the height \( h \) can be expressed as:
\[ h = \frac{1}{2} g T^2 \]

3. **Distance Fallen in \( \frac{T}{5} \) Seconds**:
We need to find the distance fallen by the ball in \( \frac{T}{5} \) seconds. Using the formula for distance:
\[ x = \frac{1}{2} g \left(\frac{T}{5}\right)^2 \]
Simplifying this gives:
\[ x = \frac{1}{2} g \cdot \frac{T^2}{25} = \frac{g T^2}{50} \]

4. **Relating \( x \) to \( h \)**:
From the expression for \( h \):
\[ h = \frac{1}{2} g T^2 \]
We can substitute \( \frac{g T^2}{2} \) for \( h \) in the equation for \( x \):
\[ x = \frac{h}{25} \]

5. **Finding the Position Above the Ground**:
The position of the ball above the ground after \( \frac{T}{5} \) seconds is given by:
\[ \text{Position above ground} = h - x \]
Substituting \( x \):
\[ \text{Position above ground} = h - \frac{h}{25} = \frac{25h}{25} - \frac{h}{25} = \frac{24h}{25} \]

### Final Answer:
The position of the ball above the ground after \( \frac{T}{5} \) seconds is:
\[ \frac{24h}{25} \text{ meters} \]