Two identical object A and B each of mass m start moving along the same vertical line in opposite directions at the same instant. Object A is dropped from rest from a height H above the ground and object B is projected vertically upward from the ground with speed = `sqrt(2gh)`.
At what height above the ground do they collide ?

Options

A. `(1)/(4) H`
B. `(1)/(2) H`
C. `(2)/(3) H`
D. `(3)/(4)H`

Explanation

To solve the problem, we need to determine the height at which two objects, A and B, collide when object A is dropped from a height \( H \) and object B is projected upwards from the ground with a speed of \( \sqrt{2gH} \).

### Step-by-Step Solution:

1. **Understand the Motion of Object A**:
- Object A is dropped from rest from a height \( H \).
- The initial velocity \( u_A = 0 \).
- The distance fallen by object A after time \( t \) is given by the equation of motion:
\[ h_A = H - \frac{1}{2} g t^2 \]

2. **Understand the Motion of Object B**:
- Object B is projected upwards with an initial speed \( u_B = \sqrt{2gH} \).
- The distance traveled upwards by object B after time \( t \) is given by:
\[ h_B = u_B t - \frac{1}{2} g t^2 = \sqrt{2gH} t - \frac{1}{2} g t^2 \]

3. **Set Up the Collision Condition**:
- The objects collide when they are at the same height, i.e., \( h_A = h_B \).
- Therefore, we set the equations equal to each other:
\[ H - \frac{1}{2} g t^2 = \sqrt{2gH} t - \frac{1}{2} g t^2 \]

4. **Simplify the Equation**:
- Cancel \( -\frac{1}{2} g t^2 \) from both sides:
\[ H = \sqrt{2gH} t \]

5. **Solve for Time \( t \)**:
- Rearranging gives:
\[ t = \frac{H}{\sqrt{2gH}} = \frac{\sqrt{H}}{\sqrt{2g}} = \sqrt{\frac{H}{2g}} \]

6. **Find the Height of Collision**:
- Substitute \( t \) back into the equation for \( h_A \) (or \( h_B \)):
\[ h = H - \frac{1}{2} g t^2 \]
- Substitute \( t^2 = \frac{H}{2g} \):
\[ h = H - \frac{1}{2} g \left(\frac{H}{2g}\right) = H - \frac{H}{4} = \frac{3H}{4} \]

### Final Answer:
The height at which the two objects collide is:
\[ \boxed{\frac{3H}{4}} \]