Drops of water fall from the roof of a building 9m. High at regular intervals of time, the first drop reaching the ground at the same instant fourth drop starts to fall. What are the distance of the second and third drops from the roof?

Options

A. 4m and 1 m
B. 4m and 2 m
C. 6m and 2 m
D. 6m and 3 m

Answer

To solve the problem, we need to find the distances of the second and third drops from the roof of a building that is 9 meters high. The drops fall at regular intervals of time, and the first drop reaches the ground at the same instant the fourth drop starts to fall.

### Step-by-Step Solution:

1. **Understanding the Time Interval**:
Let's denote the time interval between the drops as \( t \). The first drop reaches the ground after a time \( T \), and the fourth drop starts falling at that instant. Since there are three drops before the fourth drop, the time taken for the first drop to reach the ground is \( T = 3t \).

2. **Using the Equation of Motion**:
The height \( h \) from which the drops fall can be described using the equation of motion:
\[ h = ut + \frac{1}{2} a t^2 \]
Here, \( u = 0 \) (initial velocity), \( a = g \) (acceleration due to gravity), and \( h = 9 \) m (height of the building). Therefore, the equation simplifies to:
\[ 9 = \frac{1}{2} g T^2 \]

3. **Substituting \( T \)**:
Since \( T = 3t \), we substitute this into the equation:
\[ 9 = \frac{1}{2} g (3t)^2 \]
This simplifies to:
\[ 9 = \frac{1}{2} g \cdot 9t^2 \]
\[ 9 = \frac{9}{2} g t^2 \]
Dividing both sides by 9 gives:
\[ 1 = \frac{1}{2} g t^2 \]
Hence,
\[ t^2 = \frac{2}{g} \]

4. **Calculating \( t \)**:
Assuming \( g \approx 10 \, \text{m/s}^2 \):
\[ t^2 = \frac{2}{10} = \frac{1}{5} \]
Therefore,
\[ t = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \, \text{s} \]

5. **Finding the Distance of the Second Drop**:
The second drop falls after \( t \) seconds. Using the equation of motion:
\[ h_2 = \frac{1}{2} g t^2 \]
Substituting \( t^2 = \frac{1}{5} \):
\[ h_2 = \frac{1}{2} \cdot 10 \cdot \frac{1}{5} = 1 \, \text{m} \]
Thus, the distance of the second drop from the roof is:
\[ 9 - h_2 = 9 - 1 = 8 \, \text{m} \]

6. **Finding the Distance of the Third Drop**:
The third drop falls after \( 2t \) seconds. Using the equation of motion:
\[ h_3 = \frac{1}{2} g (2t)^2 \]
\[ h_3 = \frac{1}{2} g \cdot 4t^2 = 2g t^2 \]
Substituting \( t^2 = \frac{1}{5} \):
\[ h_3 = 2 \cdot 10 \cdot \frac{1}{5} = 4 \, \text{m} \]
Thus, the distance of the third drop from the roof is:
\[ 9 - h_3 = 9 - 4 = 5 \, \text{m} \]

### Final Answers:
- The distance of the second drop from the roof is **8 meters**.
- The distance of the third drop from the roof is **5 meters**.