A body is falling from height 'h' it takes `t_(1)` time to rech the ground. The time taken to cover the first half of height is:-

Options

A. `t_(2)=(t_(1))/sqrt(2)`
B. `t_(1)=(t_(2))/sqrt(2)`
C. `t_(2)=sqrt(3)t_(1)`
D. None of these

Explanation

To solve the problem of finding the time taken to cover the first half of the height when a body is falling from a height \( h \), we can use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body is falling from a height \( h \) and takes time \( t_1 \) to reach the ground.
- We need to find the time \( t_2 \) taken to cover the first half of the height, which is \( \frac{h}{2} \).

2. **Using the Equation of Motion**:
- The equation for the distance fallen under gravity is given by:
\[ h = \frac{1}{2} g t_1^2 \]
- Here, \( g \) is the acceleration due to gravity.

3. **Finding the Time for Half the Height**:
- For the first half of the height \( \frac{h}{2} \), we can use the same equation:
\[ \frac{h}{2} = \frac{1}{2} g t_2^2 \]
- This simplifies to:
\[ h = g t_2^2 \]

4. **Setting Up the Equations**:
- From the first equation, we can express \( h \) in terms of \( t_1 \):
\[ h = \frac{1}{2} g t_1^2 \]
- From the second equation, we have:
\[ h = g t_2^2 \]

5. **Equating the Two Expressions for \( h \)**:
- Set the two expressions for \( h \) equal to each other:
\[ \frac{1}{2} g t_1^2 = g t_2^2 \]
- We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \frac{1}{2} t_1^2 = t_2^2 \]

6. **Solving for \( t_2 \)**:
- Rearranging gives:
\[ t_2^2 = \frac{1}{2} t_1^2 \]
- Taking the square root of both sides:
\[ t_2 = \frac{t_1}{\sqrt{2}} \]

7. **Final Answer**:
- Therefore, the time taken to cover the first half of the height is:
\[ t_2 = \frac{t_1}{\sqrt{2}} \]