A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time `(t)/(2)` sec?

Options

A. At `(h)/(2)` from the ground
B. At `(h)/(4)` from the ground
C. depends upon mass and volume of the body
D. at `(3h)/(4)` from the ground

Explanation

To solve the problem step by step, we will use the equations of motion under gravity.

### Step 1: Understand the problem
A body is released from the top of a tower of height \( h \) and takes \( t \) seconds to reach the ground. We need to find the position of the body after \( \frac{t}{2} \) seconds.

### Step 2: Use the equation of motion
The equation of motion for an object under free fall (with initial velocity \( u = 0 \)) is given by:
\[ h = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \) (the initial velocity), \( a = g \) (acceleration due to gravity), and \( t = t \) (the total time taken to reach the ground). Therefore, the equation simplifies to:
\[ h = 0 \cdot t + \frac{1}{2} g t^2 \]
This can be rewritten as:
\[ h = \frac{1}{2} g t^2 \tag{1} \]

### Step 3: Find the distance covered in \( \frac{t}{2} \) seconds
Now, we need to find the distance \( h' \) covered in \( \frac{t}{2} \) seconds. We can use the same equation of motion:
\[ h' = u \left(\frac{t}{2}\right) + \frac{1}{2} g \left(\frac{t}{2}\right)^2 \]
Substituting \( u = 0 \):
\[ h' = 0 + \frac{1}{2} g \left(\frac{t}{2}\right)^2 \]
This simplifies to:
\[ h' = \frac{1}{2} g \cdot \frac{t^2}{4} = \frac{1}{8} g t^2 \tag{2} \]

### Step 4: Relate \( h' \) to \( h \)
From equation (1), we know that \( h = \frac{1}{2} g t^2 \). We can express \( g t^2 \) in terms of \( h \):
\[ g t^2 = 2h \]
Substituting this into equation (2):
\[ h' = \frac{1}{8} \cdot 2h = \frac{h}{4} \]

### Step 5: Determine the position of the ball
After \( \frac{t}{2} \) seconds, the ball has fallen \( \frac{h}{4} \) from the top of the tower. Therefore, the distance from the ground is:
\[ \text{Distance from the ground} = h - h' = h - \frac{h}{4} = \frac{3h}{4} \]

### Final Answer
The ball will be at a height of \( \frac{3h}{4} \) from the ground after \( \frac{t}{2} \) seconds.

---