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

Options

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

Answer

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 \( h \) and takes time \( t_1 \) to reach the ground, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Motion**:
The body is falling freely under gravity, starting from rest. The initial velocity \( u = 0 \), the acceleration \( a = g \), and the total distance fallen \( s = h \) after time \( t_1 \).

2. **Using the Second Equation of Motion**:
The second equation of motion states:
\[ s = ut + \frac{1}{2} a t^2 \]
For the full height \( h \):
\[ h = 0 \cdot t_1 + \frac{1}{2} g t_1^2 \]
This simplifies to:
\[ h = \frac{1}{2} g t_1^2 \quad \text{(Equation 1)} \]

3. **Finding the Time for Half the Height**:
We need to find the time \( t_2 \) taken to fall half the height \( \frac{h}{2} \):
\[ \frac{h}{2} = 0 \cdot t_2 + \frac{1}{2} g t_2^2 \]
This simplifies to:
\[ \frac{h}{2} = \frac{1}{2} g t_2^2 \quad \text{(Equation 2)} \]

4. **Relating the Two Equations**:
From Equation 1, we have:
\[ h = \frac{1}{2} g t_1^2 \]
Substituting this into Equation 2 gives:
\[ \frac{1}{2} \left(\frac{1}{2} g t_1^2\right) = \frac{1}{2} g t_2^2 \]
Simplifying this, we get:
\[ \frac{1}{2} g t_1^2 = g t_2^2 \]

5. **Canceling Common Terms**:
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ \frac{1}{2} t_1^2 = t_2^2 \]

6. **Taking the Square Root**:
Taking the square root of both sides gives:
\[ t_2 = \frac{t_1}{\sqrt{2}} \]

### Final Answer:
The time taken to cover the first half of the height is:
\[ t_2 = \frac{t_1}{\sqrt{2}} \]