A projectile of range R bursts at its highest point in two fragments, Both pieces at the time of bursts have the velocity in the horizontal direction. The heavier is double the mass of the figther. Lighter frangments falls at `(R)/(2)` horizontal distance from the point of projection in the opposite side of projection. The distance. where other part falls, from point of projection.

Options

A. `(7R)/(4)`
B. `(5R)/(4)`
C. `(8R)/(4)`
D. `(6R)/(4)`

Explanation

To solve the problem, we will analyze the motion of the projectile and the effects of the explosion at its highest point.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A projectile is launched and reaches a maximum height before bursting into two fragments.
- The heavier fragment has double the mass of the lighter fragment.
- The lighter fragment lands at a distance of \( \frac{R}{2} \) from the point of projection in the opposite direction.

2. **Defining Masses**:
- Let the mass of the lighter fragment be \( m \).
- Then, the mass of the heavier fragment will be \( 2m \).

3. **Conservation of Momentum**:
- At the moment of the burst, the horizontal momentum before the burst must equal the total horizontal momentum after the burst.
- Before the burst, the horizontal momentum is given by:
\[ P_{\text{initial}} = (m + 2m) \cdot v_x = 3m \cdot v_x \]
- After the burst, the lighter fragment moves left (opposite direction) with a velocity \( v \), and the heavier fragment moves right with a velocity \( V \):
\[ P_{\text{final}} = m(-v) + 2mV \]

4. **Setting Up the Equation**:
- By conservation of momentum:
\[ 3m \cdot v_x = m(-v) + 2mV \]
- Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ 3v_x = -v + 2V \]

5. **Finding the Distance for the Heavier Fragment**:
- The lighter fragment travels \( \frac{R}{2} \) to the left.
- Let \( x \) be the distance traveled by the heavier fragment to the right.
- Since the total horizontal distance covered by both fragments must equal the range \( R \):
\[ R = x + \frac{R}{2} \]
- Rearranging gives:
\[ x = R - \frac{R}{2} = \frac{R}{2} \]

6. **Calculating the Total Distance**:
- The distance from the point of projection to where the heavier fragment lands is:
\[ \text{Distance} = \frac{R}{2} + R = \frac{3R}{2} \]

### Final Answer:
The distance where the heavier fragment falls from the point of projection is \( \frac{3R}{2} \).