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. `h(2 - m//M)`
C. 2 h
D. a function of `m,h and u`.

Explanation

To solve the problem, we need to analyze the situation in a gravity-free space where a man throws a ball downwards. We will use the concept of conservation of momentum and the center of mass.

### Step-by-Step Solution:

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 \( u \).

2. **Initial Conditions**:
Before the ball is thrown, the system (man + ball) is at rest. The total momentum of the system is zero.

3. **Throwing the Ball**:
When the man throws the ball downwards, he exerts a force on the ball, which results in the ball moving downwards. By Newton's third law, the ball exerts an equal and opposite force on the man, causing him to move upwards.

4. **Conservation of Momentum**:
Since there are no external forces acting on the system, the total momentum before and after the throw must remain constant. Initially, the momentum is zero:
\[ 0 = M v_{\text{man}} + m (-u) \]
Here, \( v_{\text{man}} \) is the velocity of the man moving upwards after throwing the ball.

5. **Solving for the Man's Velocity**:
Rearranging the equation gives:
\[ M v_{\text{man}} = m u \]
Thus, the velocity of the man after throwing the ball is:
\[ v_{\text{man}} = \frac{m u}{M} \]

6. **Finding the Distance Moved by the Man**:
The distance the man moves upwards while the ball falls can be calculated using the time it takes for the ball to reach the floor. The ball falls a distance \( h \) with an initial speed \( u \). Using the equation of motion:
\[ h = u t + \frac{1}{2} (0) t^2 \quad \text{(since acceleration is zero)} \]
This simplifies to:
\[ h = u t \implies t = \frac{h}{u} \]

7. **Distance Moved by the Man**:
The distance \( d \) that the man moves upwards in time \( t \) is given by:
\[ d = v_{\text{man}} \cdot t = \left(\frac{m u}{M}\right) \cdot \left(\frac{h}{u}\right) = \frac{m h}{M} \]

8. **Final Height of the Man Above the Floor**:
The new height of the man above the floor when the ball reaches the floor is:
\[ h' = h + d = h + \frac{m h}{M} = h \left(1 + \frac{m}{M}\right) \]

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