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

Explanation

To solve the problem, we need to determine the distances of the second and third drops from the roof of a building that is 9 meters high. The first drop reaches the ground at the same time the fourth drop starts to fall. Let's break it down step by step.

### Step 1: Understand the Timing of the Drops
Since drops are falling at regular intervals, let’s denote the time interval between each drop as \( t \). Therefore:
- The first drop falls for \( t_1 = T \) (the total time taken to reach the ground).
- The second drop falls for \( t_2 = T - t \).
- The third drop falls for \( t_3 = T - 2t \).
- The fourth drop starts falling at \( t_4 = T \).

### Step 2: Calculate the Time Taken by the First Drop
Using the equation of motion for the first drop:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s = 9 \, \text{m} \) (height of the building),
- \( u = 0 \) (initial velocity),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

Substituting the values:
\[ 9 = 0 \cdot T + \frac{1}{2} \cdot 10 \cdot T^2 \]
This simplifies to:
\[ 9 = 5T^2 \implies T^2 = \frac{9}{5} \implies T = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}} \, \text{s} \]

### Step 3: Calculate the Distances of the Second and Third Drops
Now, we can calculate the distances of the second and third drops from the roof.

#### Distance of the Second Drop
The second drop falls for \( t_2 = T - t \):
\[ t_2 = T - t = T - \frac{T}{3} = \frac{2T}{3} \]
Now, substituting \( t_2 \) into the equation of motion:
\[ s_2 = \frac{1}{2} g t_2^2 = \frac{1}{2} \cdot 10 \cdot \left(\frac{2T}{3}\right)^2 \]
Calculating \( s_2 \):
\[ s_2 = 5 \cdot \frac{4T^2}{9} = \frac{20T^2}{9} \]
Substituting \( T^2 = \frac{9}{5} \):
\[ s_2 = \frac{20 \cdot \frac{9}{5}}{9} = 4 \, \text{m} \]
Thus, the distance of the second drop from the roof is \( 4 \, \text{m} \).

#### Distance of the Third Drop
The third drop falls for \( t_3 = T - 2t \):
\[ t_3 = T - 2t = T - \frac{2T}{3} = \frac{T}{3} \]
Now, substituting \( t_3 \) into the equation of motion:
\[ s_3 = \frac{1}{2} g t_3^2 = \frac{1}{2} \cdot 10 \cdot \left(\frac{T}{3}\right)^2 \]
Calculating \( s_3 \):
\[ s_3 = 5 \cdot \frac{T^2}{9} = \frac{5T^2}{9} \]
Substituting \( T^2 = \frac{9}{5} \):
\[ s_3 = \frac{5 \cdot \frac{9}{5}}{9} = 1 \, \text{m} \]
Thus, the distance of the third drop from the roof is \( 1 \, \text{m} \).

### Final Result
- Distance of the second drop from the roof: **4 m**
- Distance of the third drop from the roof: **1 m**