The balls are released from the top of a tower of heigh `H` at regular interval of time. When first ball reaches at the grund, the nthe ball is to be just released and `((n+1))/(2) th`
ball is at some distance `h` from top of the tower. Find the value of `h`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
We have a tower of height \( H \) from which balls are dropped at regular time intervals. When the first ball reaches the ground, the \( n \)-th ball is just released, and the \( \frac{n+1}{2} \)-th ball is at a height \( h \) from the top of the tower.

### Step 2: Determine the time intervals
Let \( t \) be the time interval between the releases of the balls. Therefore:
- The first ball is dropped at \( t = 0 \).
- The second ball is dropped at \( t = t \).
- The third ball is dropped at \( t = 2t \).
- The \( n \)-th ball is dropped at \( t = (n-1)t \).

### Step 3: Calculate the time taken for the first ball to reach the ground
The first ball falls from height \( H \) under the influence of gravity. The distance fallen by the first ball when it reaches the ground can be described by the equation:
\[ H = \frac{1}{2} g t_1^2 \]
where \( t_1 \) is the time taken for the first ball to reach the ground.

### Step 4: Relate \( t_1 \) to \( n \)
Since the \( n \)-th ball is released just as the first ball reaches the ground, we have:
\[ t_1 = (n-1)t \]
Substituting this into the equation for \( H \):
\[ H = \frac{1}{2} g ((n-1)t)^2 \]
This simplifies to:
\[ H = \frac{1}{2} g (n-1)^2 t^2 \]

### Step 5: Determine the height of the \( \frac{n+1}{2} \)-th ball
The time taken for the \( \frac{n+1}{2} \)-th ball to fall from the top of the tower is:
\[ t_2 = \left(\frac{n+1}{2} - 1\right)t = \left(\frac{n-1}{2}\right)t \]
The distance fallen by the \( \frac{n+1}{2} \)-th ball when the first ball reaches the ground is:
\[ h = H - \frac{1}{2} g t_2^2 \]
Substituting \( t_2 \):
\[ h = H - \frac{1}{2} g \left(\frac{n-1}{2}t\right)^2 \]
This simplifies to:
\[ h = H - \frac{1}{2} g \frac{(n-1)^2}{4} t^2 \]
\[ h = H - \frac{1}{8} g (n-1)^2 t^2 \]

### Step 6: Substitute \( H \) from Step 4 into the equation for \( h \)
From Step 4, we know:
\[ H = \frac{1}{2} g (n-1)^2 t^2 \]
Substituting this into the equation for \( h \):
\[ h = \frac{1}{2} g (n-1)^2 t^2 - \frac{1}{8} g (n-1)^2 t^2 \]
Factoring out \( g (n-1)^2 t^2 \):
\[ h = g (n-1)^2 t^2 \left(\frac{1}{2} - \frac{1}{8}\right) \]
Calculating the fraction:
\[ \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8} \]
Thus:
\[ h = \frac{3}{8} g (n-1)^2 t^2 \]

### Step 7: Final expression for \( h \)
Since we know that \( H = \frac{1}{2} g (n-1)^2 t^2 \), we can express \( h \) in terms of \( H \):
\[ h = \frac{3}{4} H \]

### Conclusion
The height \( h \) from the top of the tower where the \( \frac{n+1}{2} \)-th ball is located is:
\[ h = \frac{3H}{4} \]