A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in `T/3` second?

Options

A. `h/9`metre from the ground
B. `(7h//9)` metre from the ground
C. `(8h//9)` metre from the ground
D. `(17h//18)` metre from the ground

Explanation

To solve the problem, we need to find the position of the ball after \( \frac{T}{3} \) seconds when it is released from a height \( H \) meters.

### Step-by-step Solution:

1. **Understanding the Motion**:
The ball is released from rest, so its initial velocity \( u = 0 \). The only force acting on the ball is gravity, which accelerates it downwards with an acceleration \( g \).

2. **Using the Equation of Motion**:
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s \) is the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

3. **Substituting Values for Total Time \( T \)**:
Since the ball takes \( T \) seconds to reach the ground, we can set up the equation for the total distance \( H \):
\[ H = 0 \cdot T + \frac{1}{2} g T^2 \]
This simplifies to:
\[ H = \frac{1}{2} g T^2 \]

4. **Finding Distance Traveled in \( \frac{T}{3} \) Seconds**:
Now, we need to find the distance traveled in \( \frac{T}{3} \) seconds. We substitute \( t = \frac{T}{3} \) into the equation:
\[ s = 0 \cdot \frac{T}{3} + \frac{1}{2} g \left(\frac{T}{3}\right)^2 \]
This simplifies to:
\[ s = \frac{1}{2} g \cdot \frac{T^2}{9} = \frac{g T^2}{18} \]

5. **Relating \( s \) to \( H \)**:
From our earlier equation \( H = \frac{1}{2} g T^2 \), we can express \( g T^2 \) in terms of \( H \):
\[ g T^2 = 2H \]
Therefore,
\[ s = \frac{2H}{18} = \frac{H}{9} \]

6. **Finding the Position from the Ground**:
The distance \( s \) is the distance fallen from the top of the tower. To find the position of the ball from the ground, we subtract the distance fallen from the total height:
\[ \text{Position from ground} = H - s = H - \frac{H}{9} = \frac{9H}{9} - \frac{H}{9} = \frac{8H}{9} \]

### Final Answer:
The position of the ball from the ground after \( \frac{T}{3} \) seconds is:
\[ \frac{8H}{9} \text{ meters from the ground.} \]