In the climax of a movie, the hero jumps from a helicopter and the villain chasing the hero also jumps from the same level . After sometime when they come at same horizontal level, the villain fires bullet horizontally towards the hero, Both were falling with constant acceleration `2(m)/(s^2), because of parachute, Assuming the hero to be within the range of bullet, which of the following is correct.

Options

A. bullet will hit the hero
B. bullet will pass above the hero
C. bullet will pass below the hero
D. bullet will definitely hit the hero, if both were falling with constant acceleration `g` instead of `2(m)/(s^2)`

Explanation

To solve the problem, we need to analyze the motion of both the hero and the villain, as well as the bullet fired by the villain.

### Step-by-step Solution:

1. **Identify the Initial Conditions**:
- Both the hero and the villain jump from the same height and fall with a constant downward acceleration of \(2 \, \text{m/s}^2\) due to their parachutes.

2. **Acceleration of the Bullet**:
- When the villain fires the bullet, it will experience a downward acceleration due to gravity, which is approximately \(g = 10 \, \text{m/s}^2\).

3. **Relative Motion**:
- Since both the hero and the villain are falling with an acceleration of \(2 \, \text{m/s}^2\), we can say that they are both in free fall but with different accelerations compared to the bullet.

4. **Time of Flight**:
- When the villain fires the bullet horizontally, it will start falling downwards with an acceleration of \(10 \, \text{m/s}^2\). The bullet will fall faster than the hero, who is falling at \(2 \, \text{m/s}^2\).

5. **Comparison of Positions**:
- As time progresses, the vertical position of the hero can be described by the equation:
\[ y_h = h - \frac{1}{2} \cdot 2 \cdot t^2 \]
- The vertical position of the bullet can be described by:
\[ y_b = h - \frac{1}{2} \cdot 10 \cdot t^2 \]
- Here, \(h\) is the initial height from which both the hero and the bullet start falling.

6. **Finding the Difference in Positions**:
- The difference in the vertical positions of the bullet and the hero at any time \(t\) can be calculated as:
\[ y_b - y_h = \left(h - \frac{1}{2} \cdot 10 \cdot t^2\right) - \left(h - \frac{1}{2} \cdot 2 \cdot t^2\right) \]
- Simplifying this gives:
\[ y_b - y_h = \frac{1}{2} \cdot (10 - 2) \cdot t^2 = \frac{1}{2} \cdot 8 \cdot t^2 = 4t^2 \]
- This means that the bullet is falling \(4t^2\) meters below the hero at any time \(t\).

7. **Conclusion**:
- Since the bullet falls faster than the hero, it will pass below the hero when fired. Therefore, the correct conclusion is that the bullet will not hit the hero, as it will fall below him.

### Final Answer:
The bullet will pass below the hero.