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 of the projectile that bursts into two fragments at its highest point, we will follow these steps:

### Step 1: Understand the Problem
We have a projectile with a range \( R \) that bursts into two fragments at its highest point. The heavier fragment has double the mass of the lighter fragment. The lighter fragment falls at a horizontal distance of \( \frac{R}{2} \) in the opposite direction from the point of projection.

### Step 2: Define the Masses
Let:
- The mass of the lighter fragment be \( m_1 \).
- The mass of the heavier fragment be \( m_2 = 2m_1 \).

The total mass of the projectile is:
\[ m = m_1 + m_2 = m_1 + 2m_1 = 3m_1 \]
Thus, we can express \( m_1 \) as:
\[ m_1 = \frac{m}{3}, \quad m_2 = \frac{2m}{3} \]

### Step 3: Center of Mass Calculation
At the moment of bursting, the center of mass of the two fragments will continue to move in the same horizontal direction as the original projectile. The position of the center of mass \( x_{cm} \) can be expressed as:
\[ x_{cm} = \frac{m_1 \cdot (-\frac{R}{2}) + m_2 \cdot (R + x)}{m_1 + m_2} \]
Where \( x \) is the distance from the point of projection where the heavier fragment lands.

### Step 4: Substitute Known Values
Substituting the values of \( m_1 \) and \( m_2 \):
\[ R = \frac{\frac{m}{3} \cdot (-\frac{R}{2}) + \frac{2m}{3} \cdot (R + x)}{m} \]

### Step 5: Simplify the Equation
Multiply through by \( m \) to eliminate the denominator:
\[ mR = \frac{m}{3} \cdot (-\frac{R}{2}) + \frac{2m}{3} \cdot (R + x) \]
This simplifies to:
\[ mR = -\frac{mR}{6} + \frac{2mR}{3} + \frac{2mx}{3} \]

### Step 6: Combine Like Terms
Rearranging gives:
\[ mR + \frac{mR}{6} = \frac{2mR}{3} + \frac{2mx}{3} \]
To combine the terms:
\[ \frac{6mR + mR}{6} = \frac{4mR + 2mx}{3} \]
This simplifies to:
\[ \frac{7mR}{6} = \frac{4mR + 2mx}{3} \]

### Step 7: Solve for \( x \)
Cross-multiplying gives:
\[ 7mR \cdot 3 = 6(4mR + 2mx) \]
This simplifies to:
\[ 21mR = 24mR + 12mx \]
Rearranging gives:
\[ 21mR - 24mR = 12mx \]
\[ -3mR = 12mx \]
Dividing by \( m \) (assuming \( m \neq 0 \)):
\[ -3R = 12x \]
Thus:
\[ x = -\frac{3R}{12} = -\frac{R}{4} \]

### Step 8: Calculate the Distance from the Origin
The distance where the heavier fragment falls from the point of projection is:
\[ R + x = R - \frac{R}{4} = \frac{3R}{4} \]

### Final Answer
The distance where the heavier fragment falls from the point of projection is \( \frac{3R}{4} \).