In a gravity free space, man of mass `M` standing at a height `h` above the floor, throws a ball of mass `m` straight down with a speed `u`. When the ball reaches the floor, the distance of the man above the floor will be.

Options

A. `h(1+(m/M))`
B. `(1+(M/m))h`
C. h
D. `(m/M)h`

Explanation

To solve the problem step by step, we will analyze the situation using the principles of the center of mass in a gravity-free environment.

### Step 1: Understand the System
We have a man of mass \( M \) standing at a height \( h \) above the floor. He throws a ball of mass \( m \) straight down with an initial speed \( u \). Since this is in gravity-free space, the only forces acting are those due to the masses of the man and the ball.

**Hint:** Visualize the setup: a man at height \( h \) and a ball being thrown downwards.

### Step 2: Center of Mass Before the Throw
Initially, the center of mass (CM) of the system (man + ball) can be calculated using the formula:
\[ \text{CM}_{\text{initial}} = \frac{M \cdot h + m \cdot 0}{M + m} = \frac{M \cdot h}{M + m} \]
Here, we consider the floor as the reference point (height = 0).

**Hint:** Remember that the center of mass is a weighted average of the positions of the masses.

### Step 3: Center of Mass After the Throw
When the man throws the ball down, the center of mass of the system must remain unchanged because there are no external forces acting on it. Let’s denote the height the man rises after throwing the ball as \( h' \).

The new position of the center of mass after the ball is thrown will be:
\[ \text{CM}_{\text{final}} = \frac{M \cdot (h + h') + m \cdot 0}{M + m} = \frac{M \cdot (h + h')}{M + m} \]

**Hint:** The center of mass remains constant in the absence of external forces.

### Step 4: Set the Center of Mass Before and After Equal
Since the center of mass does not change, we can set the initial and final center of mass equal:
\[ \frac{M \cdot h}{M + m} = \frac{M \cdot (h + h')}{M + m} \]

**Hint:** This equality reflects the conservation of the center of mass.

### Step 5: Solve for \( h' \)
From the equality of the center of mass, we can simplify:
\[ M \cdot h = M \cdot (h + h') \]
This simplifies to:
\[ M \cdot h = M \cdot h + M \cdot h' \implies 0 = M \cdot h' \implies h' = -\frac{m}{M} \cdot h \]

**Hint:** The negative sign indicates that the man moves upward when the ball is thrown down.

### Step 6: Calculate the Final Height of the Man
The final height of the man above the floor can be expressed as:
\[ \text{Height of man above floor} = h + h' = h - \frac{m}{M} \cdot h = h \left(1 + \frac{m}{M}\right) \]

**Hint:** Factor out \( h \) to simplify the expression.

### Final Answer
Thus, the distance of the man above the floor when the ball reaches the floor is:
\[ \text{Distance} = h \left(1 + \frac{m}{M}\right) \]