A body is dropped from a height H. The time taken to cover second half of the journey is

Options

A. `2sqrt((2H)/(g))`
B. `sqrt((H)/(g))`
C. `sqrt((H)/(g))(sqrt2-1)`
D. `sqrt((2H)/(g))xx(1)/((sqrt2-1))`

Answer

To solve the problem of finding the time taken to cover the second half of the journey when a body is dropped from a height \( H \), we can follow these steps:

### Step 1: Understand the Problem
A body is dropped from a height \( H \). We need to find the time taken to cover the second half of the journey, which means from \( H/2 \) to \( H \).

### Step 2: Use the Equation of Motion
The equation of motion we will use is:
\[ S = ut + \frac{1}{2} a t^2 \]
where:
- \( S \) is the distance covered,
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which is \( g \) for free fall),
- \( t \) is the time taken.

### Step 3: Total Time to Fall from Height \( H \)
For the total distance \( H \):
- Initial velocity \( u = 0 \)
- Acceleration \( a = g \)

Using the equation:
\[ H = 0 \cdot T + \frac{1}{2} g T^2 \]
This simplifies to:
\[ H = \frac{1}{2} g T^2 \]
From this, we can solve for \( T \):
\[ T^2 = \frac{2H}{g} \quad \Rightarrow \quad T = \sqrt{\frac{2H}{g}} \]

### Step 4: Time to Fall the First Half \( H/2 \)
Now, we need to find the time taken to fall the first half of the height \( H/2 \):
\[ \frac{H}{2} = 0 \cdot t + \frac{1}{2} g t^2 \]
This simplifies to:
\[ \frac{H}{2} = \frac{1}{2} g t^2 \]
From this, we can solve for \( t \):
\[ H = g t^2 \quad \Rightarrow \quad t^2 = \frac{H}{g} \quad \Rightarrow \quad t = \sqrt{\frac{H}{g}} \]

### Step 5: Time for the Second Half
The time taken to cover the second half of the journey is the total time \( T \) minus the time for the first half \( t \):
\[ t' = T - t \]
Substituting the values we found:
\[ t' = \sqrt{\frac{2H}{g}} - \sqrt{\frac{H}{g}} \]
Factoring out \( \sqrt{\frac{H}{g}} \):
\[ t' = \sqrt{\frac{H}{g}} \left( \sqrt{2} - 1 \right) \]

### Final Answer
Thus, the time taken to cover the second half of the journey is:
\[ t' = \sqrt{\frac{H}{g}} \left( \sqrt{2} - 1 \right) \]