Suppose that water drops are released from a point at the edge of a roof with a constant time interval `Deltat` between one water drop and the next. The drops fall a distance h to the ground. If `Deltat` is very short ie the number of drops falling through the air at any given instant is very large then the CM of the drops is very nearly at a height (above the ground ) of

Options

A. `h//2`
B. `h//3`
C. `2h//3`
D. `3h//4`

Explanation

To solve the problem, we need to find the height of the center of mass (CM) of the water drops that are falling from a height \( h \) with a constant time interval \( \Delta t \) between each drop.

### Step-by-Step Solution:

1. **Understanding the Motion of Drops**:
Each drop falls freely under gravity. The distance fallen by each drop after \( n \Delta t \) seconds can be calculated using the equation of motion:
\[ y_n = \frac{1}{2} g (n \Delta t)^2 \]
where \( g \) is the acceleration due to gravity.

2. **Distance Fallen by Each Drop**:
The first drop (which just started falling) will have fallen a distance of:
\[ y_1 = 0 \]
The second drop (which has fallen for \( \Delta t \)) will have fallen:
\[ y_2 = \frac{1}{2} g (\Delta t)^2 \]
The third drop (which has fallen for \( 2 \Delta t \)) will have fallen:
\[ y_3 = \frac{1}{2} g (2 \Delta t)^2 = 2g (\Delta t)^2 \]
Continuing this pattern, the \( n \)-th drop will have fallen:
\[ y_n = \frac{1}{2} g (n \Delta t)^2 \]

3. **Finding the Heights of the Drops**:
The height of each drop above the ground can be expressed as:
\[ H_n = h - y_n = h - \frac{1}{2} g (n \Delta t)^2 \]

4. **Calculating the Center of Mass**:
The center of mass \( Y_{CM} \) of the drops can be calculated using the formula:
\[ Y_{CM} = \frac{1}{N} \sum_{i=1}^{N} H_i \]
where \( N \) is the total number of drops.

5. **Summing the Heights**:
We can express the sum of the heights as:
\[ \sum_{n=1}^{N} H_n = \sum_{n=1}^{N} \left( h - \frac{1}{2} g (n \Delta t)^2 \right) = Nh - \frac{1}{2} g (\Delta t)^2 \sum_{n=1}^{N} n^2 \]
The sum of the squares of the first \( N \) natural numbers is given by:
\[ \sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6} \]

6. **Substituting the Sum**:
Substituting this into our expression for \( Y_{CM} \):
\[ Y_{CM} = \frac{1}{N} \left( Nh - \frac{1}{2} g (\Delta t)^2 \cdot \frac{N(N+1)(2N+1)}{6} \right) \]

7. **Simplifying the Expression**:
As \( N \) becomes very large, we can approximate:
\[ Y_{CM} \approx h - \frac{g (\Delta t)^2}{12} \]
This indicates that the center of mass is slightly below the original height \( h \).

8. **Final Height Above the Ground**:
The height of the center of mass above the ground is:
\[ Y_{CM} = h - \frac{h}{3} = \frac{2h}{3} \]

### Conclusion:
Thus, the height of the center of mass of the drops above the ground is:
\[ \boxed{\frac{2h}{3}} \]