A trolley is moving with velocity `v_(1)` in the horizontal direction . A bullet is fired from his trolley in the upward direction with a speed `v_(2)` . Observer is standing on the ground and observes the bullet to be moving in a parabolic path. What will be the range of projectile ?

Options

A. `(2v_(1)v_(2))/(g)`
B. `(2v_(1)^(2))/(g)`
C. `(2v_(2)^(2))/(g)`
D. `(v_(1)v_(2))/(g)`

Explanation

To solve the problem of finding the range of the projectile (the bullet) fired from a moving trolley, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The trolley is moving horizontally with a velocity \( v_1 \).
- The bullet is fired vertically upward with a speed \( v_2 \).
- An observer on the ground sees the bullet moving in a parabolic path due to the combination of its vertical and horizontal velocities.

2. **Components of Velocity**:
- The bullet has two components of velocity:
- Vertical component: \( v_2 \) (upward)
- Horizontal component: \( v_1 \) (due to the motion of the trolley)

3. **Time of Flight**:
- The time of flight \( T \) for the bullet can be calculated using the vertical motion equations.
- The bullet will go up and then come down. The time to reach the maximum height can be calculated using:
\[ T_{up} = \frac{v_2}{g} \]
- The total time of flight \( T \) (up and down) will be:
\[ T = 2 \times T_{up} = 2 \times \frac{v_2}{g} = \frac{2v_2}{g} \]

4. **Horizontal Range**:
- The horizontal range \( R \) can be calculated using the horizontal velocity and the total time of flight:
\[ R = v_1 \times T \]
- Substituting the expression for \( T \):
\[ R = v_1 \times \frac{2v_2}{g} \]
- Therefore, the range of the projectile is:
\[ R = \frac{2v_1 v_2}{g} \]

### Final Answer:
The range of the projectile (bullet) is given by:
\[ R = \frac{2v_1 v_2}{g} \]