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 m`
B. ` 7h//9 m`
C. ` 8h//9m`
D. ` 17 h//18 m`

Explanation

To solve the problem, we need to determine the position of the ball after \( T/3 \) seconds when it is released from the top of a tower of height \( h \) meters.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The ball is released from rest, so its initial velocity \( u = 0 \) m/s. It falls under the influence of gravity, which means it experiences a constant acceleration \( g \) (approximately \( 9.81 \, \text{m/s}^2 \)).

2. **Using the Equation of Motion**:
We can use the second equation of motion to find the distance traveled by the ball in time \( T \):
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s \) is the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. Substituting the known values:
\[ h = 0 \cdot T + \frac{1}{2} g T^2 \]
This simplifies to:
\[ h = \frac{1}{2} g T^2 \]

3. **Finding the Distance Traveled in \( T/3 \) Seconds**:
Now, we need to find the distance traveled by the ball in \( T/3 \) seconds. We can use the same equation of motion:
\[ s' = u \left(\frac{T}{3}\right) + \frac{1}{2} g \left(\frac{T}{3}\right)^2 \]
Substituting \( u = 0 \):
\[ s' = 0 + \frac{1}{2} g \left(\frac{T}{3}\right)^2 \]
This simplifies to:
\[ s' = \frac{1}{2} g \frac{T^2}{9} = \frac{g T^2}{18} \]

4. **Relating \( s' \) to \( h \)**:
From our earlier equation, we know that \( h = \frac{1}{2} g T^2 \). Thus, we can express \( s' \) in terms of \( h \):
\[ s' = \frac{g T^2}{18} = \frac{1}{9} \left(\frac{1}{2} g T^2\right) = \frac{h}{9} \]

5. **Finding the Position from the Ground**:
The position of the ball from the ground after \( T/3 \) seconds is:
\[ \text{Position from ground} = h - s' = h - \frac{h}{9} = \frac{9h}{9} - \frac{h}{9} = \frac{8h}{9} \]

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