A body freely falling from the rest has velocity `v` after it falls through a height `h` the distance it has to fall down for its velocity to become double is

Options

A. 2h
B. 4h
C. 6h
D. 8h

Explanation

To solve the problem, we need to determine the distance a body must fall for its velocity to double from `v` to `2v`. We will use the kinematic equations of motion under constant acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The body is falling freely from rest, so its initial velocity \( u = 0 \).
- After falling a height \( h \), its velocity becomes \( v \).

2. **Using the Kinematic Equation**:
- We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
- Here, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration (which is \( g \) for free fall), and \( s \) is the displacement (which is \( h \) in this case).
- Substituting the values:
\[ v^2 = 0 + 2gh \quad \text{(since } u = 0\text{)} \]
- Therefore, we have:
\[ v^2 = 2gh \quad \text{(Equation 1)} \]

3. **Finding the Distance for Velocity to Double**:
- Now, we want to find the distance \( x \) that the body must fall to reach a velocity of \( 2v \).
- Using the same kinematic equation for the new scenario:
\[ (2v)^2 = u^2 + 2as \]
- Here, the initial velocity \( u = v \) (the velocity at height \( h \)), and \( a = g \). The displacement is \( x \).
- Substituting the values:
\[ 4v^2 = v^2 + 2gx \]
- Rearranging gives:
\[ 4v^2 - v^2 = 2gx \]
\[ 3v^2 = 2gx \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2**:
- From Equation 1, we know \( v^2 = 2gh \). Substitute this into Equation 2:
\[ 3(2gh) = 2gx \]
- Simplifying gives:
\[ 6gh = 2gx \]
- Dividing both sides by \( 2g \) (assuming \( g \neq 0 \)):
\[ 3h = x \]
- Therefore, the distance \( x \) that the body must fall to double its velocity is:
\[ x = 3h \]

### Final Answer:
The distance the body has to fall down for its velocity to become double is \( 3h \).