A body of mass `m` is dropped from a height of `h`. Simultaneously another body of mass `2m` is thrown up vertically with such a velocity `v` that they collide at height `h/2`. If the collision is perfectly inelastic, the velocity of combined mass at the time of collision with the ground will be

Options

A. (a) `sqrt((5gh)/(4))`
B. (b) `sqrt(gh)`
C. (c) `sqrt((gh)/(4))`
D. (d) None of these

Explanation

To solve the problem step by step, we will analyze the motion of both bodies and apply the principles of conservation of momentum.

### Step 1: Determine the time taken for both bodies to collide at height \( h/2 \).

1. **For the mass \( m \)** (dropped from height \( h \)):
- Initial velocity \( u_1 = 0 \)
- Distance \( s = h/2 \)
- Acceleration \( a = g \) (acceleration due to gravity)

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ \frac{h}{2} = 0 + \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{h}{g} \implies t = \sqrt{\frac{h}{g}} \]

### Step 2: Calculate the velocity of mass \( m \) just before the collision.

Using the equation:
\[ v_1 = u_1 + at \]
Substituting the values:
\[ v_1 = 0 + g t = g \sqrt{\frac{h}{g}} = \sqrt{gh} \]

### Step 3: Determine the initial velocity \( v \) of mass \( 2m \) thrown upwards.

1. **For the mass \( 2m \)** (thrown upwards):
- Initial velocity \( u_2 = v \)
- Distance \( s = h/2 \)
- Acceleration \( a = -g \) (acting downwards)

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
\[ \frac{h}{2} = v t - \frac{1}{2} g t^2 \]
Substituting \( t = \sqrt{\frac{h}{g}} \):
\[ \frac{h}{2} = v \sqrt{\frac{h}{g}} - \frac{1}{2} g \left(\sqrt{\frac{h}{g}}\right)^2 \]
Simplifying gives:
\[ \frac{h}{2} = v \sqrt{\frac{h}{g}} - \frac{h}{2} \]
Rearranging gives:
\[ v \sqrt{\frac{h}{g}} = h \implies v = \sqrt{gh} \]

### Step 4: Calculate the velocities just before the collision.

At the time of collision:
- Velocity of mass \( m \) just before collision: \( v_1 = \sqrt{gh} \)
- Velocity of mass \( 2m \) just before collision (upwards): \( v_2 = v - gt = \sqrt{gh} - g\sqrt{\frac{h}{g}} = 0 \)

### Step 5: Apply conservation of momentum for the perfectly inelastic collision.

Using conservation of momentum:
\[ m v_1 + 2m v_2 = (m + 2m) V \]
Substituting the values:
\[ m (\sqrt{gh}) + 2m (0) = 3m V \]
This simplifies to:
\[ m \sqrt{gh} = 3m V \implies V = \frac{\sqrt{gh}}{3} \]

### Step 6: Find the velocity of the combined mass just before it hits the ground.

The combined mass will fall from height \( h/2 \) with an initial velocity \( V \):
Using the equation of motion:
\[ v^2 = u^2 + 2gh \]
Where \( u = \frac{\sqrt{gh}}{3} \) and \( h = \frac{h}{2} \):
\[ v^2 = \left(\frac{\sqrt{gh}}{3}\right)^2 + 2g\left(\frac{h}{2}\right) \]
Calculating gives:
\[ v^2 = \frac{gh}{9} + gh = \frac{gh}{9} + \frac{9gh}{9} = \frac{10gh}{9} \]
Thus, the final velocity \( v \) just before hitting the ground:
\[ v = \sqrt{\frac{10gh}{9}} = \frac{\sqrt{10gh}}{3} \]

### Final Answer:
The velocity of the combined mass at the time of collision with the ground is \( \frac{\sqrt{10gh}}{3} \).