A particle of mass m is dropped from a height h above the ground. Simultaneously another particle of the same mass is thrown vertically upwards from the ground with a speed of`sqrt(2gh)`. If they collide head-on completely inelastically, then the time taken for the combined mass to reach the ground is

Options

A. `sqrt((3h)/( 4g))`
B. `sqrt((3h)/( 2g))`
C. `sqrt((h)/(2g))`
D. `sqrt((h)/( 4g))`

Answer

To solve the problem, we need to analyze the motion of both particles and find the time taken for the combined mass to reach the ground after they collide.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Particle A (mass m) is dropped from a height \( h \).
- Particle B (mass m) is thrown upwards from the ground with an initial speed \( v_0 = \sqrt{2gh} \).

2. **Determine the Equations of Motion**:
- For Particle A (falling down):
- Initial velocity \( u_A = 0 \)
- Distance fallen after time \( t \):
\[ h' = h - \frac{1}{2} g t^2 \]
- For Particle B (moving upwards):
- Initial velocity \( u_B = \sqrt{2gh} \)
- Distance traveled upwards after time \( t \):
\[ h'' = u_B t - \frac{1}{2} g t^2 = \sqrt{2gh} t - \frac{1}{2} g t^2 \]

3. **Set Up the Collision Condition**:
- At the moment of collision, the total distance covered by both particles must equal the initial height \( h \):
\[ h' + h'' = h \]
Substituting the expressions for \( h' \) and \( h'' \):
\[ \left( h - \frac{1}{2} g t^2 \right) + \left( \sqrt{2gh} t - \frac{1}{2} g t^2 \right) = h \]
Simplifying this gives:
\[ h - g t^2 + \sqrt{2gh} t = h \]
Thus,
\[ \sqrt{2gh} t - g t^2 = 0 \]
Factoring out \( t \):
\[ t (\sqrt{2gh} - gt) = 0 \]
This gives us two solutions: \( t = 0 \) (initial time) or \( t = \frac{\sqrt{2gh}}{g} \).

4. **Calculate the Time of Collision**:
- The time of collision is:
\[ t = \frac{\sqrt{2gh}}{g} = \sqrt{\frac{2h}{g}} \]

5. **Determine the Height of the Combined Mass After Collision**:
- After the collision, the two particles stick together and fall as a single mass. The height of the combined mass at the moment of collision can be calculated using either particle's position at time \( t \).

6. **Calculate the Height of the Combined Mass**:
- Using Particle A's height:
\[ h' = h - \frac{1}{2} g t^2 = h - \frac{1}{2} g \left(\frac{2h}{g}\right) = h - h = 0 \]
- The combined mass starts falling from the height where they collide.

7. **Calculate the Time Taken to Reach the Ground**:
- The combined mass falls from a height of \( \frac{3h}{4} \) (as derived from the equations) to the ground. The time taken to fall this distance is given by:
\[ t' = \sqrt{\frac{2h'}{g}} = \sqrt{\frac{2 \cdot \frac{3h}{4}}{g}} = \sqrt{\frac{3h}{2g}} \]

8. **Total Time Taken**:
- The total time taken for the combined mass to reach the ground is:
\[ T = t + t' = \sqrt{\frac{2h}{g}} + \sqrt{\frac{3h}{2g}} = \sqrt{\frac{3h}{2g}} \]

### Final Answer:
The time taken for the combined mass to reach the ground after the collision is:
\[ T = \sqrt{\frac{3h}{2g}} \]