A body A is thrown up vertically from the ground with velocity `V_(0)` and another body B is simultaneously dropped from a height H. They meet simultaneously dropped from a height H. They meet at a height `(H)/(2)` if `V_(0)` is equal to

Options

A. `sqrt(2gH)`
B. `sqrt(gH)`
C. `(1)/(2)sqrt(gH)`
D. `sqrt((2g)/(H))`

Explanation

To solve the problem, we need to find the initial velocity \( V_0 \) of body A, which is thrown upwards from the ground, such that it meets body B, which is dropped from a height \( H \), at a height of \( \frac{H}{2} \).

### Step-by-Step Solution:

1. **Understand the Motion of Body A (Thrown Upwards)**:
- Body A is thrown upwards with an initial velocity \( V_0 \).
- The height \( y_A \) of body A after time \( t \) can be expressed using the equation of motion:
\[ y_A = V_0 t - \frac{1}{2} g t^2 \]
- Here, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

2. **Understand the Motion of Body B (Dropped Downwards)**:
- Body B is dropped from a height \( H \) with an initial velocity of \( 0 \).
- The height \( y_B \) of body B after time \( t \) can be expressed as:
\[ y_B = H - \frac{1}{2} g t^2 \]

3. **Set Up the Equation for Meeting Point**:
- The bodies meet at a height of \( \frac{H}{2} \). Therefore, we can set the heights equal:
\[ y_A = y_B = \frac{H}{2} \]

4. **Substituting for Body A**:
- Substitute \( y_A \) into the equation:
\[ \frac{H}{2} = V_0 t - \frac{1}{2} g t^2 \quad \text{(1)} \]

5. **Substituting for Body B**:
- Substitute \( y_B \) into the equation:
\[ \frac{H}{2} = H - \frac{1}{2} g t^2 \]
- Rearranging gives:
\[ \frac{1}{2} g t^2 = H - \frac{H}{2} = \frac{H}{2} \]
- Thus, we have:
\[ g t^2 = H \quad \text{(2)} \]

6. **Solving for Time \( t \)**:
- From equation (2):
\[ t^2 = \frac{H}{g} \quad \Rightarrow \quad t = \sqrt{\frac{H}{g}} \]

7. **Substituting \( t \) back into Equation (1)**:
- Substitute \( t \) into equation (1):
\[ \frac{H}{2} = V_0 \sqrt{\frac{H}{g}} - \frac{1}{2} g \left(\sqrt{\frac{H}{g}}\right)^2 \]
- Simplifying gives:
\[ \frac{H}{2} = V_0 \sqrt{\frac{H}{g}} - \frac{1}{2} g \cdot \frac{H}{g} \]
- This simplifies to:
\[ \frac{H}{2} = V_0 \sqrt{\frac{H}{g}} - \frac{H}{2} \]

8. **Rearranging to Solve for \( V_0 \)**:
- Adding \( \frac{H}{2} \) to both sides:
\[ H = V_0 \sqrt{\frac{H}{g}} \]
- Dividing both sides by \( \sqrt{\frac{H}{g}} \):
\[ V_0 = \frac{H}{\sqrt{\frac{H}{g}}} = \sqrt{gH} \]

### Final Answer:
The value of \( V_0 \) is:
\[ V_0 = \sqrt{gH} \, \text{m/s} \]