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 behavior of the center of mass after the explosion. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
The bomb is projected from the ground with an initial speed \( v \) at an angle \( \theta \) with the horizontal. The initial velocity can be resolved into horizontal and vertical components:
- Horizontal component: \( v_x = v \cos(\theta) \)
- Vertical component: \( v_y = v \sin(\theta) \)

### Step 2: Determine the Time to Reach Maximum Height
The time \( t \) to reach maximum height can be found using the vertical motion equation. At maximum height, the vertical velocity is zero:
\[ v_y - g t = 0 \implies t = \frac{v \sin(\theta)}{g} \]

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

### Step 4: Analyze the Explosion
At maximum height, the bomb explodes into two fragments of equal mass \( \frac{m}{2} \). One fragment comes to rest immediately after the explosion, while the other continues moving.

### Step 5: Center of Mass Motion
Since no external horizontal forces act on the system, the center of mass of the system will continue to move with the same horizontal velocity it had just before the explosion. The horizontal velocity of the center of mass \( v_{cm} \) is:
\[ v_{cm} = v \cos(\theta) \]

### Step 6: Calculate the Horizontal Range of the Center of Mass
The horizontal range \( R \) of the center of mass can be calculated using the time of flight for the entire projectile motion. The total time of flight \( T \) is:
\[ T = 2t = 2 \left(\frac{v \sin(\theta)}{g}\right) = \frac{2v \sin(\theta)}{g} \]
The horizontal range \( R \) is given by:
\[ R = v_{cm} \cdot T = (v \cos(\theta)) \left(\frac{2v \sin(\theta)}{g}\right) = \frac{2v^2 \sin(\theta) \cos(\theta)}{g} \]
Using the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ R = \frac{v^2 \sin(2\theta)}{g} \]

### Conclusion
The horizontal range of the center of mass after the explosion is:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]