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

Explanation

To find the position of the ball after \( \frac{T}{3} \) seconds when it is released from the top of a tower of height \( h \) meters, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**: The ball is released from rest, meaning its initial velocity \( u = 0 \). The only acceleration acting on the ball is due to gravity, denoted as \( g \).

2. **Use the Equation of Motion**: The equation of motion that relates distance, initial velocity, acceleration, and time is:
\[ s = ut + \frac{1}{2} a t^2 \]
For our scenario, since the ball is released from rest, we have:
\[ s = 0 + \frac{1}{2} g t^2 \]
Therefore, the distance \( h \) (the height of the tower) when the ball reaches the ground after time \( T \) is:
\[ h = \frac{1}{2} g T^2 \]

3. **Find the Position at \( \frac{T}{3} \)**: We need to find the position of the ball after \( \frac{T}{3} \) seconds. We can use the same equation of motion:
\[ s_1 = u \left(\frac{T}{3}\right) + \frac{1}{2} g \left(\frac{T}{3}\right)^2 \]
Substituting \( u = 0 \):
\[ s_1 = 0 + \frac{1}{2} g \left(\frac{T}{3}\right)^2 = \frac{1}{2} g \cdot \frac{T^2}{9} = \frac{g T^2}{18} \]

4. **Relate \( s_1 \) to \( h \)**: From the earlier step, we know \( h = \frac{1}{2} g T^2 \). Therefore, we can express \( s_1 \) in terms of \( h \):
\[ s_1 = \frac{g T^2}{18} = \frac{1}{9} \cdot \frac{1}{2} g T^2 = \frac{h}{9} \]

5. **Calculate the Position from the Ground**: The distance covered by the ball from the top of the tower after \( \frac{T}{3} \) seconds is \( \frac{h}{9} \). Thus, the height of the ball from the ground is:
\[ \text{Height from the ground} = h - s_1 = h - \frac{h}{9} = \frac{8h}{9} \]

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