`B_(1)`,`B_(2)`, and `B_(3)`, are three balloos ascending with velocities `v`, `2v`, and `3v`, respectively, If a bomb is dropped from each when they are at the same height, then.

Options

A. Bomb from `B_(1)` reaches ground first
B. Bomb from `B_(2)` reaches ground first
C. Bomb from `B_(3)` reaches ground first
D. They reach the ground simultaneously

Explanation

To solve the problem, we need to analyze the motion of the bombs dropped from the three balloons, B1, B2, and B3, which are ascending with velocities v, 2v, and 3v respectively.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The balloons are at the same height when the bombs are dropped.
- The velocities of the balloons are:
- Balloon B1: \( v_1 = v \)
- Balloon B2: \( v_2 = 2v \)
- Balloon B3: \( v_3 = 3v \)

2. **Analyzing the Motion of the Bombs**:
- When a bomb is dropped from a balloon, it initially has the same upward velocity as the balloon.
- The bombs will experience the acceleration due to gravity (g) acting downwards.

3. **Finding the Maximum Height**:
- Each bomb will first ascend to a maximum height before descending.
- The time taken to reach the maximum height can be calculated using the formula:
\[ t = \frac{u}{g} \]
- Where \( u \) is the initial velocity of the bomb (which is the same as the velocity of the balloon at the moment of release).

4. **Calculating the Maximum Height for Each Bomb**:
- For bomb from B1 (initial velocity = v):
\[ h_1 = \frac{v^2}{2g} \]
- For bomb from B2 (initial velocity = 2v):
\[ h_2 = \frac{(2v)^2}{2g} = \frac{4v^2}{2g} = \frac{2v^2}{g} \]
- For bomb from B3 (initial velocity = 3v):
\[ h_3 = \frac{(3v)^2}{2g} = \frac{9v^2}{2g} \]

5. **Total Distance Fallen by Each Bomb**:
- After reaching the maximum height, each bomb will fall to the ground. The total distance fallen for each bomb is the sum of the height it ascends and the distance it falls back to the ground.
- The time taken to fall from height \( h \) to the ground can be calculated using:
\[ t = \sqrt{\frac{2h}{g}} \]

6. **Calculating Total Time for Each Bomb**:
- For bomb from B1:
\[ t_1 = \frac{v}{g} + \sqrt{\frac{2h_1}{g}} = \frac{v}{g} + \sqrt{\frac{2 \cdot \frac{v^2}{2g}}{g}} = \frac{v}{g} + \sqrt{\frac{v^2}{g^2}} = \frac{v}{g} + \frac{v}{g} = \frac{2v}{g} \]
- For bomb from B2:
\[ t_2 = \frac{2v}{g} + \sqrt{\frac{2h_2}{g}} = \frac{2v}{g} + \sqrt{\frac{2 \cdot \frac{2v^2}{g}}{g}} = \frac{2v}{g} + \sqrt{\frac{4v^2}{g^2}} = \frac{2v}{g} + \frac{2v}{g} = \frac{4v}{g} \]
- For bomb from B3:
\[ t_3 = \frac{3v}{g} + \sqrt{\frac{2h_3}{g}} = \frac{3v}{g} + \sqrt{\frac{2 \cdot \frac{9v^2}{2g}}{g}} = \frac{3v}{g} + \sqrt{\frac{9v^2}{g^2}} = \frac{3v}{g} + \frac{3v}{g} = \frac{6v}{g} \]

7. **Comparing Times**:
- From the calculations:
- \( t_1 = \frac{2v}{g} \)
- \( t_2 = \frac{4v}{g} \)
- \( t_3 = \frac{6v}{g} \)
- The bomb from B1 reaches the ground first because it takes the least time.

### Conclusion:
The bomb from balloon B1 reaches the ground first.