A body is released from the top of a tower of height h and takes 'T' sec to reach the ground. The position of the body at T/2 sec is

Options

A. `h/4`from ground
B. `(3h)/4` from ground
C. `h/8` from ground
D. `(5h)/4` from ground

Answer

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

### Step-by-Step Solution:

1. **Understanding the Problem:**
- A body is released from the top of a tower of height \( h \).
- It takes time \( T \) seconds to reach the ground.
- We need to find the position of the body at \( \frac{T}{2} \) seconds.

2. **Using the Second Equation of Motion:**
- The second equation of motion is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s \) is the distance fallen, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

3. **Initial Conditions:**
- Since the body is released (not thrown), the initial velocity \( u = 0 \).
- The acceleration \( a = g \) (acceleration due to gravity).

4. **Finding Time of Flight:**
- The total distance \( s \) when the body reaches the ground is equal to the height of the tower \( h \):
\[ h = 0 \cdot T + \frac{1}{2} g T^2 \]
- Simplifying this gives:
\[ h = \frac{1}{2} g T^2 \]
- Rearranging for \( T \):
\[ T = \sqrt{\frac{2h}{g}} \]

5. **Finding Position at \( \frac{T}{2} \):**
- Now, we need to find the position of the body at \( t = \frac{T}{2} \):
\[ s = 0 \cdot \frac{T}{2} + \frac{1}{2} g \left(\frac{T}{2}\right)^2 \]
- Substituting \( \frac{T}{2} \):
\[ s = \frac{1}{2} g \left(\frac{T^2}{4}\right) = \frac{g T^2}{8} \]

6. **Substituting for \( T^2 \):**
- From the earlier equation, we know \( T^2 = \frac{2h}{g} \):
\[ s = \frac{g}{8} \cdot \frac{2h}{g} = \frac{h}{4} \]

7. **Finding the Position from the Ground:**
- The distance fallen \( s \) is \( \frac{h}{4} \). Therefore, the height from the ground is:
\[ \text{Height from the ground} = h - s = h - \frac{h}{4} = \frac{3h}{4} \]

### Final Answer:
The position of the body at \( \frac{T}{2} \) seconds is \( \frac{3h}{4} \) from the ground.