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

Options

A. `(sqrt40-2)/6`
B. `(sqrt30-3)/7`
C. `(sqrt20-2)/5`
D. `sqrt(3/2)`

Explanation

To solve the problem step by step, we will analyze the motion of both particles and their collision.

### Step 1: Analyze the motion of the first particle (mass m)
The first particle is dropped from a height \( h \). The distance it falls after time \( t_0 \) can be calculated using the second equation of motion:
\[ H_1 = \frac{1}{2} g t_0^2 \]
where \( g \) is the acceleration due to gravity.

### Step 2: Analyze the motion of the second particle (mass 2m)
The second particle is thrown upwards from the ground with an initial speed \( v_0 = \sqrt{gh} \). The distance it travels upwards after time \( t_0 \) is given by:
\[ H_2 = v_0 t_0 - \frac{1}{2} g t_0^2 \]
Substituting \( v_0 \):
\[ H_2 = \sqrt{gh} \cdot t_0 - \frac{1}{2} g t_0^2 \]

### Step 3: Set up the equation for collision
At the time of collision, the sum of the distances traveled by both particles must equal the height \( h \):
\[ H_1 + H_2 = h \]
Substituting the expressions for \( H_1 \) and \( H_2 \):
\[ \frac{1}{2} g t_0^2 + \left(\sqrt{gh} \cdot t_0 - \frac{1}{2} g t_0^2\right) = h \]
This simplifies to:
\[ \sqrt{gh} \cdot t_0 = h \]
From this, we can solve for \( t_0 \):
\[ t_0 = \frac{h}{\sqrt{gh}} = \frac{\sqrt{h}}{\sqrt{g}} = \sqrt{\frac{h}{g}} \]

### Step 4: Calculate the velocities just before the collision
For the first particle (mass \( m \)):
Using the first equation of motion:
\[ v_1 = g t_0 = g \sqrt{\frac{h}{g}} = \sqrt{gh} \]

For the second particle (mass \( 2m \)):
Using the first equation of motion:
\[ v_2 = v_0 - g t_0 = \sqrt{gh} - g \sqrt{\frac{h}{g}} = 0 \]

### Step 5: Calculate the velocity after collision
In a completely inelastic collision, the momentum before collision equals the momentum after collision:
\[ m v_1 + 2m v_2 = 3m v' \]
Substituting the values:
\[ m (\sqrt{gh}) + 2m (0) = 3m v' \]
This simplifies to:
\[ \sqrt{gh} = 3v' \implies v' = \frac{\sqrt{gh}}{3} \]

### Step 6: Calculate the time taken to reach the ground after collision
The combined mass (3m) will move upwards with velocity \( v' \) and then fall back down. The time taken to reach the maximum height can be calculated using:
\[ t_{\text{up}} = \frac{v'}{g} = \frac{\frac{\sqrt{gh}}{3}}{g} = \frac{\sqrt{h}}{3\sqrt{g}} \]
The maximum height reached after collision is:
\[ h' = \frac{(v')^2}{2g} = \frac{\left(\frac{\sqrt{gh}}{3}\right)^2}{2g} = \frac{gh}{18g} = \frac{h}{18} \]

### Step 7: Calculate total height from which the mass falls
The total height from which the combined mass falls is:
\[ \text{Total height} = \frac{h}{2} + \frac{h}{18} = \frac{9h}{18} + \frac{h}{18} = \frac{10h}{18} = \frac{5h}{9} \]

### Step 8: Calculate the time taken to fall from this height
Using the second equation of motion:
\[ h = \frac{1}{2} g t^2 \implies t^2 = \frac{2h}{g} \implies t = \sqrt{\frac{2h}{g}} \]

### Step 9: Total time taken
The total time taken for the combined mass to reach the ground is:
\[ t_{\text{total}} = t_{\text{up}} + t_{\text{fall}} = \frac{\sqrt{h}}{3\sqrt{g}} + \sqrt{\frac{2h}{g}} = \sqrt{\frac{h}{g}} \left(\frac{1}{3} + \sqrt{2}\right) \]

### Final Answer
In units of \( \sqrt{\frac{h}{g}} \), the time taken for the combined mass to reach the ground is:
\[ \frac{1}{3} + \sqrt{2} \]