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 determine the time taken for the combined mass to reach the ground after they collide.

### Step 1: Determine the time taken for both particles to collide

1. **Particle A (dropped from height h)**:
- Initial velocity (u) = 0
- Distance fallen = h
- Using the second equation of motion:
\[ h = ut + \frac{1}{2}gt^2 \implies h = 0 + \frac{1}{2}gt^2 \implies t^2 = \frac{2h}{g} \implies t = \sqrt{\frac{2h}{g}} \]

2. **Particle B (thrown upwards with speed \(\sqrt{2gh}\))**:
- Initial velocity (u) = \(\sqrt{2gh}\)
- Distance traveled upwards = h
- Using the second equation of motion:
\[ h = ut - \frac{1}{2}gt^2 \implies h = \sqrt{2gh}t - \frac{1}{2}gt^2 \]
- Rearranging gives:
\[ \frac{1}{2}gt^2 - \sqrt{2gh}t + h = 0 \]

### Step 2: Solve the quadratic equation for time

Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

- Here, \(a = \frac{1}{2}g\), \(b = -\sqrt{2gh}\), and \(c = -h\).
- The discriminant \(D\) is:
\[ D = b^2 - 4ac = (\sqrt{2gh})^2 - 4 \cdot \frac{1}{2}g \cdot (-h) = 2gh + 2gh = 4gh \]
- Thus, the time \(t\) becomes:
\[ t = \frac{\sqrt{2gh} \pm \sqrt{4gh}}{g} = \frac{\sqrt{2gh} \pm 2\sqrt{gh}}{g} \]
- We take the positive root since time cannot be negative:
\[ t = \frac{(1 + 2)\sqrt{gh}}{g} = \frac{3\sqrt{gh}}{g} \]

### Step 3: Determine the height at which they collide

Using the time \(t\) in the equation for Particle A:
\[ h' = \frac{1}{2}g\left(\frac{3\sqrt{h}}{\sqrt{g}}\right)^2 = \frac{1}{2}g \cdot \frac{9h}{g} = \frac{9h}{2} \]

### Step 4: Calculate the time taken for the combined mass to reach the ground after collision

After the collision, the combined mass will fall from the height \(h' = \frac{3h}{4}\) with an initial velocity of 0. The time taken to fall this height is given by:
\[ t_{fall} = \sqrt{\frac{2h'}{g}} = \sqrt{\frac{2 \cdot \frac{3h}{4}}{g}} = \sqrt{\frac{3h}{2g}} \]

### Final Answer

The total time taken for the combined mass to reach the ground after collision is:
\[ t_{total} = t + t_{fall} = \sqrt{\frac{2h}{g}} + \sqrt{\frac{3h}{2g}} \]