A particle of mass 2m is dropped from a height 80 m above the ground. At the same time another particle of mass m is thrown vertically upwards from the ground with a speed of 40 m/s. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of second is:

Options

A. (a)`((sqrt(112) - 2)/(3))`
B. (b)`((sqrt(80) - 2)/(3))`
C. (c)`((sqrt(105) - 2)/(5))`
D. (d)`((sqrt(112) + 2)/(5))`

Explanation

To solve the problem, we need to analyze the motion of both particles and find out when they collide, and then determine how long it takes for the combined mass to reach the ground.

### Step-by-Step Solution:

1. **Identify the motion of the first particle (mass = 2m)**:
- The particle is dropped from a height of 80 m.
- Initial velocity \( u_1 = 0 \).
- The distance fallen after time \( t_0 \) is given by the equation:
\[ h_1 = \frac{1}{2} g t_0^2 \]
- Here, \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

2. **Identify the motion of the second particle (mass = m)**:
- The particle is thrown upwards with an initial velocity of \( u_2 = 40 \, \text{m/s} \).
- The distance traveled upwards after time \( t_0 \) is given by:
\[ h_2 = u_2 t_0 - \frac{1}{2} g t_0^2 = 40 t_0 - \frac{1}{2} g t_0^2 \]

3. **Set up the equation for collision**:
- The total distance covered by both particles when they collide is equal to the initial height:
\[ h_1 + h_2 = 80 \]
- Substituting the expressions for \( h_1 \) and \( h_2 \):
\[ \frac{1}{2} g t_0^2 + (40 t_0 - \frac{1}{2} g t_0^2) = 80 \]
- Simplifying this gives:
\[ 40 t_0 = 80 \implies t_0 = 2 \, \text{s} \]

4. **Determine the velocities at the moment of collision**:
- For the first particle (mass = 2m):
\[ v_1 = g t_0 = 10 \times 2 = 20 \, \text{m/s} \, \text{(downward)} \]
- For the second particle (mass = m):
\[ v_2 = u_2 - g t_0 = 40 - 10 \times 2 = 20 \, \text{m/s} \, \text{(upward)} \]

5. **Use conservation of momentum to find the velocity after collision**:
- The momentum before collision is:
\[ p_{\text{initial}} = (2m)(20) + (m)(-20) = 40m - 20m = 20m \]
- After the collision, the combined mass is \( 3m \) and let the common velocity be \( v_f \):
\[ p_{\text{final}} = (3m)v_f \]
- Setting the initial momentum equal to the final momentum:
\[ 20m = 3m v_f \implies v_f = \frac{20}{3} \, \text{m/s} \, \text{(downward)} \]

6. **Determine the time taken for the combined mass to reach the ground**:
- The combined mass starts from the height where the collision occurred, which is \( 60 \, \text{m} \) (since \( h_1 + h_2 = 80 \)).
- Using the second equation of motion:
\[ h = v_f t + \frac{1}{2} g t^2 \]
- Here, \( h = 60 \), \( v_f = \frac{20}{3} \), and \( g = 10 \):
\[ 60 = \frac{20}{3} t + \frac{1}{2} (10) t^2 \]
- Rearranging gives:
\[ 60 = \frac{20}{3} t + 5t^2 \]
- Multiplying through by 3 to eliminate the fraction:
\[ 180 = 20t + 15t^2 \implies 15t^2 + 20t - 180 = 0 \]
- Dividing through by 5:
\[ 3t^2 + 4t - 36 = 0 \]
- Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-36)}}{2 \cdot 3} \]
\[ t = \frac{-4 \pm \sqrt{16 + 432}}{6} = \frac{-4 \pm \sqrt{448}}{6} = \frac{-4 \pm 8\sqrt{7}}{6} \]
- Taking the positive root:
\[ t = \frac{-2 + 4\sqrt{7}}{3} \]

### Final Answer:
The time taken for the combined mass to reach the ground is \( \frac{-2 + 4\sqrt{7}}{3} \) seconds.