A shell is fired vertically upwards with a velocity `v_(1)` from the deck of a ship travelling at a speed of `v_(2)` . A person on the shore observes the motion of the shell as parabola. Its horizontal range is given by:

Options

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

Answer

To solve the problem, we need to analyze the motion of the shell fired vertically upwards from the deck of a moving ship. The shell's motion can be understood as a combination of vertical and horizontal motions, leading to a parabolic trajectory as observed from the shore.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The shell is fired vertically upwards with an initial velocity \( v_1 \).
- The ship is moving horizontally with a velocity \( v_2 \).

2. **Determine the Velocity Components**:
- The velocity of the shell with respect to the ground can be expressed as:
\[ \vec{v}_{\text{shell}} = v_2 \hat{i} + v_1 \hat{j} \]
- Here, \( v_2 \) is the horizontal component (along the x-axis) and \( v_1 \) is the vertical component (along the y-axis).

3. **Analyze the Horizontal Motion**:
- The horizontal motion of the shell is uniform (no horizontal acceleration), so the horizontal displacement \( x \) after time \( t \) is given by:
\[ x = v_2 \cdot t \]

4. **Analyze the Vertical Motion**:
- The vertical motion is subject to gravitational acceleration. The vertical displacement \( y \) after time \( t \) is given by the equation of motion:
\[ y = v_1 \cdot t - \frac{1}{2} g t^2 \]
- Since the shell returns to the same vertical level from which it was fired, we set \( y = 0 \):
\[ 0 = v_1 \cdot t - \frac{1}{2} g t^2 \]

5. **Solve for Time of Flight**:
- Rearranging the equation gives:
\[ v_1 \cdot t = \frac{1}{2} g t^2 \]
- Dividing both sides by \( t \) (assuming \( t \neq 0 \)):
\[ v_1 = \frac{1}{2} g t \]
- Solving for \( t \) gives:
\[ t = \frac{2 v_1}{g} \]

6. **Substitute Time into Horizontal Displacement**:
- Substitute \( t \) back into the horizontal displacement equation:
\[ x = v_2 \cdot t = v_2 \cdot \left(\frac{2 v_1}{g}\right) \]
- Therefore, the horizontal range \( R \) of the shell is:
\[ R = \frac{2 v_1 v_2}{g} \]

### Final Answer:
The horizontal range of the shell is given by:
\[ R = \frac{2 v_1 v_2}{g} \]