A bomb is projected with speed 10 m/s at an angle `30^o` with the horizontal. In the mid-air bomb explodes in different fragments and they all move in different directions. At the time when all fragments reach at ground, the displacement of the center of mass of bomb from initial point will be (g = 10 m/`s^2`)

Options

A. `5sqrt3` m
B. 10 m
C. 5 m
D. Zero

Answer

To solve the problem, we need to determine the displacement of the center of mass of the bomb after it is projected and explodes in mid-air. Here’s a step-by-step solution:

### Step 1: Determine the initial velocity components
The bomb is projected at a speed of \(10 \, \text{m/s}\) at an angle of \(30^\circ\) with the horizontal. We can find the horizontal and vertical components of the initial velocity (\(u\)).

- Horizontal component (\(u_x\)):
\[ u_x = u \cos(30^\circ) = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]

- Vertical component (\(u_y\)):
\[ u_y = u \sin(30^\circ) = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \]

### Step 2: Calculate the time of flight
The time of flight (\(T\)) for a projectile can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
Substituting the values:
\[ T = \frac{2 \times 5}{10} = 1 \, \text{s} \]

### Step 3: Calculate the horizontal displacement
The horizontal displacement (\(R\)) can be calculated using the horizontal velocity and the time of flight:
\[ R = u_x \times T \]
Substituting the values:
\[ R = (5\sqrt{3}) \times 1 = 5\sqrt{3} \, \text{m} \]

### Step 4: Analyze the explosion
When the bomb explodes in mid-air, the fragments will move in different directions, but the center of mass of the system will continue to follow the same trajectory as it would have if the explosion did not occur. This is due to the conservation of momentum in the absence of external horizontal forces.

### Step 5: Conclusion
The displacement of the center of mass of the bomb from the initial point when all fragments reach the ground will be the same as the horizontal displacement calculated before the explosion.

Thus, the displacement of the center of mass of the bomb from the initial point is:
\[ \text{Displacement} = 5\sqrt{3} \, \text{m} \]

### Final Answer
The displacement of the center of mass of the bomb from the initial point will be \(5\sqrt{3} \, \text{m}\).

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