A body released from the top of a tower of height h takes T seconds to reach the ground. The position of the body at T/4 seconds is

Options

A. at `h/16` from the ground
B. at `h/4` from the top of the tower
C. at `(15h)/16` from the ground
D. at `(3h)/16` from the top of the tower

Answer

To solve the problem, we need to determine the position of a body released from the top of a tower of height \( h \) after \( \frac{T}{4} \) seconds, where \( T \) is the total time taken to reach the ground.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body is released from a height \( h \).
- It takes a total time \( T \) to reach the ground.
- We want to find the position of the body at \( \frac{T}{4} \) 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, \( u = 0 \) (initial velocity since the body is released), \( a = g = 10 \, \text{m/s}^2 \) (acceleration due to gravity), and \( t = \frac{T}{4} \).

3. **Calculating the Distance Fallen in \( \frac{T}{4} \) Seconds**:
- Substitute the values into the equation:
\[ s = 0 \cdot \frac{T}{4} + \frac{1}{2} \cdot 10 \cdot \left(\frac{T}{4}\right)^2 \]
- Simplifying this:
\[ s = \frac{1}{2} \cdot 10 \cdot \frac{T^2}{16} = \frac{10T^2}{32} = \frac{5T^2}{16} \]
- Thus, the distance fallen after \( \frac{T}{4} \) seconds is \( s = \frac{5T^2}{16} \).

4. **Finding the Relation Between \( h \) and \( T \)**:
- When the body reaches the ground, the total distance fallen is equal to the height \( h \):
\[ h = ut + \frac{1}{2} a T^2 \]
- Substituting \( u = 0 \) and \( a = 10 \):
\[ h = 0 \cdot T + \frac{1}{2} \cdot 10 \cdot T^2 = 5T^2 \]

5. **Finding the Position of the Body at \( \frac{T}{4} \) Seconds**:
- We have \( s = \frac{5T^2}{16} \) and \( h = 5T^2 \).
- To find the position of the body from the top of the tower, we can express it as:
\[ \text{Position from the top} = h - s = 5T^2 - \frac{5T^2}{16} \]
- Simplifying this:
\[ \text{Position from the top} = 5T^2 \left(1 - \frac{1}{16}\right) = 5T^2 \cdot \frac{15}{16} = \frac{75T^2}{16} \]

6. **Position from the Ground**:
- The position of the body from the ground is:
\[ \text{Position from the ground} = s = \frac{5T^2}{16} \]
- Therefore, the height from the ground is:
\[ h - s = h - \frac{5T^2}{16} = 5T^2 - \frac{5T^2}{16} = \frac{15h}{16} \]

### Final Answer:
The position of the body at \( \frac{T}{4} \) seconds from the ground is \( \frac{15h}{16} \).