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 at a height `(H)/(2)` if `V_(0)` is equal to

Options

A. `sqrt(150), 12`
B. `sqrt(160), 20`
C. `sqrt(200), 30`
D. `sqrt(150), 20`

Explanation

To solve the problem, we need to analyze the motion of both bodies A and B and use the equations of motion to find the initial velocity \( V_0 \) of body A when they meet at a height \( \frac{H}{2} \).

### Step-by-Step Solution

1. **Identify the motion of body B**:
Body B is dropped from a height \( H \) with an initial velocity of \( 0 \). The distance it travels when they meet at height \( \frac{H}{2} \) is:
\[ s_B = H - \frac{H}{2} = \frac{H}{2} \]

2. **Use the second equation of motion for body B**:
The second equation of motion states:
\[ s = ut + \frac{1}{2} a t^2 \]
For body B:
- Initial velocity \( u = 0 \)
- Acceleration \( a = g \) (downward)
- Distance \( s = -\frac{H}{2} \) (since we take downward as negative)

Plugging in the values:
\[ -\frac{H}{2} = 0 \cdot t_B + \frac{1}{2} (-g) t_B^2 \]
Simplifying this gives:
\[ -\frac{H}{2} = -\frac{1}{2} g t_B^2 \]
Rearranging, we find:
\[ g t_B^2 = H \quad \Rightarrow \quad t_B^2 = \frac{H}{g} \]
Therefore:
\[ t_B = \sqrt{\frac{H}{g}} \]

3. **Identify the motion of body A**:
Body A is thrown upward with an initial velocity \( V_0 \). The distance it travels when they meet at height \( \frac{H}{2} \) is:
\[ s_A = \frac{H}{2} \]

4. **Use the second equation of motion for body A**:
For body A:
- Initial velocity \( u = V_0 \)
- Acceleration \( a = -g \) (upward motion against gravity)
- Distance \( s = \frac{H}{2} \)

Plugging in the values:
\[ \frac{H}{2} = V_0 t_A - \frac{1}{2} g t_A^2 \]

5. **Equate the time of both bodies**:
Since both bodies meet at the same time, we have \( t_A = t_B \):
\[ t_A = \sqrt{\frac{H}{g}} \]

6. **Substituting \( t_A \) into the equation for body A**:
Substitute \( t_A \) into the equation for body A:
\[ \frac{H}{2} = V_0 \left(\sqrt{\frac{H}{g}}\right) - \frac{1}{2} g \left(\sqrt{\frac{H}{g}}\right)^2 \]
Simplifying the second term:
\[ \frac{H}{2} = V_0 \sqrt{\frac{H}{g}} - \frac{1}{2} g \cdot \frac{H}{g} \]
\[ \frac{H}{2} = V_0 \sqrt{\frac{H}{g}} - \frac{H}{2} \]

7. **Rearranging to solve for \( V_0 \)**:
Adding \( \frac{H}{2} \) to both sides:
\[ H = V_0 \sqrt{\frac{H}{g}} \]
Now, solving for \( V_0 \):
\[ V_0 = \frac{H}{\sqrt{\frac{H}{g}}} \]
\[ V_0 = \sqrt{gH} \]

### Final Answer:
\[ V_0 = \sqrt{gH} \]