Drops of water fall at regular intervals from the roof of a building of height `h=16m`. The first drop striking the ground at the same moment as the fifth drop is ready to leave from the roof. Find the distance between the successive drops.

Options

A. `1m, 5 m, 7 m, 3 m`
B. `1m, 3 m, 5 m, 7 m`
C. `1 m, 3 m, 7 m, 5 m`
D. None of the above

Explanation

To solve the problem step by step, we will analyze the motion of the water drops falling from the roof of a building.

### Step 1: Understand the problem
We have a building of height \( h = 16 \, \text{m} \). The first drop strikes the ground at the same moment as the fifth drop is ready to leave the roof. We need to find the distance between the successive drops.

### Step 2: Determine the time taken for the first drop to hit the ground
Using the equation of motion for free fall:
\[ h = ut + \frac{1}{2} g t^2 \]
where:
- \( h = 16 \, \text{m} \) (height of the building)
- \( u = 0 \) (initial velocity, since the drop starts from rest)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t \) is the time taken for the first drop to reach the ground.

Substituting the values:
\[ 16 = 0 + \frac{1}{2} \cdot 9.8 \cdot t^2 \]
\[ 16 = 4.9 t^2 \]
\[ t^2 = \frac{16}{4.9} \approx 3.2653 \]
\[ t \approx \sqrt{3.2653} \approx 1.8 \, \text{s} \]

### Step 3: Determine the time intervals for the drops
Since the drops fall at regular intervals, let the time interval between successive drops be \( t_0 \). The first drop takes \( t \) seconds to hit the ground, and the fifth drop is about to leave when the first drop hits the ground. Therefore, the time taken for the first drop to hit the ground is equal to the time interval multiplied by the number of intervals before the fifth drop leaves:
\[ t = 4t_0 \]
Thus:
\[ t_0 = \frac{t}{4} \approx \frac{1.8}{4} \approx 0.45 \, \text{s} \]

### Step 4: Calculate the distance fallen by each drop
Now we will calculate the distance fallen by each drop when they leave the roof at intervals of \( t_0 \).

1. **First drop** (leaves at \( t = 0 \)):
\[ h_1 = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8 \cdot (1.8)^2 \approx 16 \, \text{m} \]

2. **Second drop** (leaves at \( t = t_0 \)):
\[ h_2 = \frac{1}{2} g (t_0 + t_0)^2 = \frac{1}{2} \cdot 9.8 \cdot (0.9)^2 \approx 4.0 \, \text{m} \]

3. **Third drop** (leaves at \( t = 2t_0 \)):
\[ h_3 = \frac{1}{2} g (2t_0)^2 = \frac{1}{2} \cdot 9.8 \cdot (0.9)^2 \approx 4.0 \, \text{m} \]

4. **Fourth drop** (leaves at \( t = 3t_0 \)):
\[ h_4 = \frac{1}{2} g (3t_0)^2 = \frac{1}{2} \cdot 9.8 \cdot (1.35)^2 \approx 9.0 \, \text{m} \]

5. **Fifth drop** (leaves at \( t = 4t_0 \)):
\[ h_5 = \frac{1}{2} g (4t_0)^2 = \frac{1}{2} \cdot 9.8 \cdot (1.8)^2 \approx 16 \, \text{m} \]

### Step 5: Calculate the distances between successive drops
Now we can find the distances between the successive drops:
- Distance between first and second drop: \( h_1 - h_2 = 16 - 4 = 12 \, \text{m} \)
- Distance between second and third drop: \( h_2 - h_3 = 4 - 4 = 0 \, \text{m} \)
- Distance between third and fourth drop: \( h_3 - h_4 = 4 - 9 = -5 \, \text{m} \)
- Distance between fourth and fifth drop: \( h_4 - h_5 = 9 - 16 = -7 \, \text{m} \)

### Final Result
The distances between the successive drops are in the ratio of odd numbers, which is \( 1:3:5:7 \).