There are two identical particles A and B. One is projected vertically upward with speed `sqrt(2gh)` from ground and other is dropped from height h along the same vertical line. Collision between them is perfectly inelastic. Find time taken by them to reach the ground after collision in terms of `sqrt(h/g)`.

Options

A. `sqrt(3/2)`
B. `sqrt(1/2)`
C. `sqrt(3)`
D. `sqrt(1/5)`

Explanation

To solve the problem, we need to analyze the motion of two particles A and B before and after their collision. Here's a step-by-step breakdown of the solution:

### Step 1: Analyze the motion of particle A
Particle A is projected vertically upward with an initial speed of \( v_A = \sqrt{2gh} \). The motion of A can be described by the following kinematic equation:

\[ y_A = v_A t - \frac{1}{2} g t^2 \]

Where:
- \( y_A \) is the height of particle A above the ground at time \( t \).
- \( g \) is the acceleration due to gravity.

### Step 2: Analyze the motion of particle B
Particle B is dropped from a height \( h \) with an initial speed of \( v_B = 0 \). The motion of B can be described by:

\[ y_B = h - \frac{1}{2} g t^2 \]

Where:
- \( y_B \) is the height of particle B above the ground at time \( t \).

### Step 3: Find the time of collision
The particles collide when their heights are equal, i.e., \( y_A = y_B \). Setting the equations equal gives:

\[ \sqrt{2gh} t - \frac{1}{2} g t^2 = h - \frac{1}{2} g t^2 \]

This simplifies to:

\[ \sqrt{2gh} t = h \]

From this, we can solve for \( t \):

\[ t = \frac{h}{\sqrt{2gh}} = \frac{\sqrt{h}}{\sqrt{2g}} \]

### Step 4: Determine the height at which the collision occurs
Substituting \( t \) back into the equation for \( y_A \) or \( y_B \) to find the height at which they collide:

Using \( y_A \):

\[ y_A = \sqrt{2gh} \left(\frac{\sqrt{h}}{\sqrt{2g}}\right) - \frac{1}{2} g \left(\frac{\sqrt{h}}{\sqrt{2g}}\right)^2 \]

Calculating this gives:

\[ y_A = \sqrt{2gh} \cdot \frac{\sqrt{h}}{\sqrt{2g}} - \frac{1}{2} g \cdot \frac{h}{2g} = h - \frac{h}{4} = \frac{3h}{4} \]

### Step 5: Find the velocity after collision
Since the collision is perfectly inelastic, both particles move together after the collision. To find their velocity just after the collision, we use the conservation of momentum:

\[ m v_A + m v_B = (m + m) v_f \]

Where:
- \( v_A \) is the velocity of A just before the collision.
- \( v_B \) is the velocity of B just before the collision.
- \( v_f \) is the final velocity after the collision.

Calculating \( v_A \) and \( v_B \) just before the collision:

1. For particle A:
\[ v_A = \sqrt{2g \left(\frac{3h}{4}\right)} = \sqrt{\frac{3gh}{2}} \]

2. For particle B:
\[ v_B = \sqrt{2gh} - g t = \sqrt{2gh} - g \cdot \frac{\sqrt{h}}{\sqrt{2g}} = \sqrt{2gh} - \frac{gh}{\sqrt{2g}} = \sqrt{2gh} - \frac{g\sqrt{h}}{\sqrt{2}} \]

Using conservation of momentum:

\[ m \sqrt{\frac{3gh}{2}} + m \left(\sqrt{2gh} - \frac{g\sqrt{h}}{\sqrt{2}}\right) = 2m v_f \]

This gives us the final velocity \( v_f \).

### Step 6: Calculate the time to reach the ground after collision
After the collision, both particles fall from a height of \( \frac{3h}{4} \) with an initial velocity \( v_f \). The time \( t' \) taken to reach the ground can be calculated using:

\[ h' = v_f t' + \frac{1}{2} g t'^2 \]

Where \( h' = \frac{3h}{4} \).

### Final Result
After solving the equations, we find that the time taken by both particles to reach the ground after the collision is:

\[ t' = \sqrt{\frac{3h}{2g}} \]