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

### Step 1: Understand the motion of the ball
The ball is released from rest, so its initial velocity \( u = 0 \). It is subjected to gravitational acceleration \( g \) downwards. The time taken to reach the ground is \( T \) seconds.

### Step 2: Use the equation of motion
The distance fallen by the ball in time \( T \) seconds can be expressed using the second equation of motion:
\[ H = uT + \frac{1}{2}gT^2 \]
Since the initial velocity \( u = 0 \), this simplifies to:
\[ H = \frac{1}{2}gT^2 \quad \text{(Equation 1)} \]

### Step 3: Find the position of the ball at \( T/3 \) seconds
Now, we need to find the position of the ball after \( T/3 \) seconds. We can use the same equation of motion:
\[ x = u\left(\frac{T}{3}\right) + \frac{1}{2}g\left(\frac{T}{3}\right)^2 \]
Again, since \( u = 0 \), this simplifies to:
\[ x = \frac{1}{2}g\left(\frac{T}{3}\right)^2 \]
Calculating \( \left(\frac{T}{3}\right)^2 \):
\[ \left(\frac{T}{3}\right)^2 = \frac{T^2}{9} \]
Thus, we have:
\[ x = \frac{1}{2}g\left(\frac{T^2}{9}\right) = \frac{gT^2}{18} \quad \text{(Equation 2)} \]

### Step 4: Relate \( gT^2 \) to \( H \)
From Equation 1, we know:
\[ gT^2 = \frac{2H}{1} \]
Substituting this into Equation 2 gives:
\[ x = \frac{1}{18} \cdot 2H = \frac{H}{9} \]

### Step 5: Find the height from the ground
The height of the ball from the ground after \( T/3 \) seconds is:
\[ \text{Height from ground} = H - x = H - \frac{H}{9} \]
Calculating this gives:
\[ H - \frac{H}{9} = \frac{9H}{9} - \frac{H}{9} = \frac{8H}{9} \]

### Final Answer
Therefore, the position of the ball from the ground after \( T/3 \) seconds is:
\[ \frac{8H}{9} \]

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