The balls are released from the top of a tower of height H at regular interval of time. When first ball reaches at the ground, the `n^(th)` ball is to be just released and `((n+1)/(2))^(th)` ball is at same distance 'h' from top of the tower. The value of h is

Options

A. `(2)/(3)H`
B. `(1)/(4)H`
C. `(4)/(5)H`
D. `(5H)/(6)`

Answer

To solve the problem step by step, we need to analyze the motion of the balls released from the top of the tower.

### Given:
- Height of the tower = H
- Balls are released at regular intervals of time (t).
- When the first ball reaches the ground, the n-th ball is just released.
- The \((n+1)/2\)-th ball is at a distance \(h\) from the top of the tower.

### Step 1: Determine the time taken for the first ball to reach the ground.
The first ball is dropped from a height \(H\) and falls freely under gravity. The equation of motion for the first ball is given by:
\[ s = ut + \frac{1}{2}gt^2 \]
Where:
- \(s = H\) (the distance fallen)
- \(u = 0\) (initial velocity, since it's dropped)
- \(g\) is the acceleration due to gravity
- \(t_1\) is the time taken for the first ball to reach the ground.

Substituting the values:
\[ H = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \]
This simplifies to:
\[ H = \frac{1}{2} g t_1^2 \]
From this, we can solve for \(t_1\):
\[ t_1^2 = \frac{2H}{g} \quad \Rightarrow \quad t_1 = \sqrt{\frac{2H}{g}} \]

### Step 2: Determine the time when the n-th ball is released.
The n-th ball is released just when the first ball reaches the ground, which means:
\[ t_n = t_1 = \sqrt{\frac{2H}{g}} \]

### Step 3: Determine the time for the \((n+1)/2\)-th ball.
The \((n+1)/2\)-th ball is released at:
\[ t_{(n+1)/2} = \left(\frac{n+1}{2} - 1\right)t = \frac{n-1}{2}t \]

### Step 4: Calculate the distance \(h\) for the \((n+1)/2\)-th ball.
Using the equation of motion for the \((n+1)/2\)-th ball:
\[ h = ut + \frac{1}{2}g t^2 \]
Where \(t = \frac{n-1}{2}t\):
\[ h = 0 \cdot \frac{n-1}{2}t + \frac{1}{2}g \left(\frac{n-1}{2}t\right)^2 \]
This simplifies to:
\[ h = \frac{1}{2}g \cdot \frac{(n-1)^2}{4}t^2 = \frac{g(n-1)^2t^2}{8} \]

### Step 5: Relate \(h\) to \(H\).
We know from the first ball's equation that:
\[ H = \frac{1}{2}g t_1^2 \]
Substituting \(t_1\) from Step 1:
\[ H = \frac{1}{2}g \cdot \frac{2H}{g} = H \]
Now, substituting \(t^2\) from \(t_1^2 = \frac{2H}{g}\):
\[ h = \frac{g(n-1)^2}{8} \cdot \frac{2H}{g} = \frac{(n-1)^2H}{4} \]

### Step 6: Find the value of \(h\).
From the previous step, we can express \(h\) as:
\[ h = \frac{H}{4} \]

### Conclusion:
Thus, the value of \(h\) is:
\[ \boxed{\frac{H}{4}} \]