A bomb of mass m is projected from the ground with speed v at angle `theta` with the horizontal. At the maximum height from the ground it explodes into two fragments of equal mass. If one fragment comes to rest immediately after explosion, then the horizontal range of centre of mass is

Options

A. `(v^(2)sin^(2)theta)/(g)`
B. `(v^(2)sin theta)/(g)`
C. `(v^(2)sin theta)/(2g)`
D. `(v^(2)sin 2theta)/(g)`

Answer

To solve the problem, we need to analyze the motion of the bomb and the subsequent explosion into two fragments. Here's a step-by-step solution:

### Step 1: Determine the initial conditions of the bomb
The bomb is projected with an initial speed \( v \) at an angle \( \theta \) with the horizontal. We can break down the initial velocity into horizontal and vertical components:
- Horizontal component of velocity, \( v_x = v \cos \theta \)
- Vertical component of velocity, \( v_y = v \sin \theta \)

### Step 2: Calculate the time to reach maximum height
At maximum height, the vertical component of the velocity becomes zero. We can use the following kinematic equation:
\[ v_y = u_y - g t \]
Setting \( v_y = 0 \) at maximum height, we have:
\[ 0 = v \sin \theta - g t \]
Solving for \( t \):
\[ t = \frac{v \sin \theta}{g} \]

### Step 3: Calculate the maximum height
The maximum height \( H \) can be calculated using the formula:
\[ H = u_y t - \frac{1}{2} g t^2 \]
Substituting \( u_y = v \sin \theta \) and \( t = \frac{v \sin \theta}{g} \):
\[ H = (v \sin \theta) \left(\frac{v \sin \theta}{g}\right) - \frac{1}{2} g \left(\frac{v \sin \theta}{g}\right)^2 \]
Simplifying this gives:
\[ H = \frac{(v \sin \theta)^2}{g} - \frac{1}{2} g \cdot \frac{(v \sin \theta)^2}{g^2} = \frac{(v \sin \theta)^2}{g} - \frac{(v \sin \theta)^2}{2g} = \frac{(v \sin \theta)^2}{2g} \]

### Step 4: Analyze the explosion at maximum height
At maximum height, the bomb explodes into two fragments of equal mass \( \frac{m}{2} \). One fragment comes to rest immediately, while the other continues moving horizontally with the same horizontal velocity \( v_x = v \cos \theta \).

### Step 5: Determine the motion of the center of mass
The center of mass of the system will continue to move horizontally with the same horizontal velocity it had before the explosion, since no external horizontal forces are acting on it. The horizontal range \( R \) of the bomb before the explosion can be calculated using the range formula:
\[ R = \frac{v^2 \sin 2\theta}{g} \]

### Step 6: Calculate the horizontal range of the center of mass
Since the center of mass continues to move with the same horizontal velocity, the horizontal range of the center of mass after the explosion will also be equal to the original range \( R \):
\[ R_{cm} = R = \frac{v^2 \sin 2\theta}{g} \]

### Final Answer
Thus, the horizontal range of the center of mass after the explosion is:
\[ R_{cm} = \frac{v^2 \sin 2\theta}{g} \]