A stone falls freely under gravity. It covered distances `h_1, h_2` and `h_3` in the first `5` seconds. The next `5` seconds and the next `5` seconds respectively. The relation between `h_1, h_2` and `h_3` is :

Options

A. `h_1 = 2h_2 = 3h_3`
B. `h_1 = (h_2)/(3) = (h_3)/(5)`
C. `h_2 = 3h_1` and `h_3 = 3h_2`
D. `h_1= h_2 = h_3`

Explanation

To solve the problem, we need to find the distances \( h_1 \), \( h_2 \), and \( h_3 \) covered by a stone falling freely under gravity during three consecutive intervals of 5 seconds each. We will use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The stone falls freely under gravity, which means its initial velocity \( u = 0 \).
- The acceleration due to gravity \( g \) is approximately \( 10 \, \text{m/s}^2 \).

2. **Distance Covered in the First 5 Seconds**:
- We can use the equation of motion:
\[ h_1 = ut + \frac{1}{2} g t^2 \]
- Substituting \( u = 0 \), \( g = 10 \, \text{m/s}^2 \), and \( t = 5 \, \text{s} \):
\[ h_1 = 0 + \frac{1}{2} \times 10 \times (5)^2 = \frac{1}{2} \times 10 \times 25 = 125 \, \text{m} \]

3. **Distance Covered in the Next 5 Seconds (5 to 10 seconds)**:
- The total time at this point is \( 10 \, \text{s} \). We calculate the total distance covered in 10 seconds:
\[ h_1 + h_2 = \frac{1}{2} g (10)^2 = \frac{1}{2} \times 10 \times 100 = 500 \, \text{m} \]
- We already found \( h_1 = 125 \, \text{m} \), so:
\[ h_2 = 500 - 125 = 375 \, \text{m} \]

4. **Distance Covered in the Next 5 Seconds (10 to 15 seconds)**:
- The total time at this point is \( 15 \, \text{s} \). We calculate the total distance covered in 15 seconds:
\[ h_1 + h_2 + h_3 = \frac{1}{2} g (15)^2 = \frac{1}{2} \times 10 \times 225 = 1125 \, \text{m} \]
- We already found \( h_1 + h_2 = 500 \, \text{m} \), so:
\[ h_3 = 1125 - 500 = 625 \, \text{m} \]

5. **Finding the Relation Between \( h_1 \), \( h_2 \), and \( h_3 \)**:
- We have:
\[ h_1 = 125 \, \text{m}, \quad h_2 = 375 \, \text{m}, \quad h_3 = 625 \, \text{m} \]
- Now, we can express the relations:
\[ h_2 = 3h_1 \quad \text{(since } 375 = 3 \times 125\text{)} \]
\[ h_3 = 5h_1 \quad \text{(since } 625 = 5 \times 125\text{)} \]

### Final Relations:
- The relations between the distances are:
\[ h_1 = \frac{h_2}{3} \quad \text{and} \quad h_1 = \frac{h_3}{5} \]