The distance moved by a freely falling body (startibg from rest) during ` st, 2nd, 3nd,……nth` second of its motion are propotional to .

Options

A. a. Even numbers
B. b. Odd numbers
C. c. All integral numbers
D. d.Squares of integral numbere

Explanation

To solve the problem of finding the distances moved by a freely falling body during the 1st, 2nd, 3rd, ..., nth seconds of its motion, we can use the equations of motion under uniform acceleration due to gravity. Here’s a step-by-step solution:

### Step 1: Understand the motion
A freely falling body starts from rest, which means its initial velocity \( u = 0 \). The acceleration \( a \) is equal to \( g \) (acceleration due to gravity), approximately \( 9.8 \, \text{m/s}^2 \).

### Step 2: Use the formula for distance covered in nth second
The distance covered during the nth second, \( s_n \), can be calculated using the formula:
\[ s_n = u + \frac{1}{2} a(2n - 1) \]
Since \( u = 0 \), this simplifies to:
\[ s_n = \frac{1}{2} g (2n - 1) \]

### Step 3: Calculate distances for the first few seconds
1. **For the 1st second (n=1)**:
\[ s_1 = \frac{1}{2} g (2 \cdot 1 - 1) = \frac{1}{2} g (1) = \frac{g}{2} \]

2. **For the 2nd second (n=2)**:
\[ s_2 = \frac{1}{2} g (2 \cdot 2 - 1) = \frac{1}{2} g (3) \]

3. **For the 3rd second (n=3)**:
\[ s_3 = \frac{1}{2} g (2 \cdot 3 - 1) = \frac{1}{2} g (5) \]

4. **For the nth second**:
\[ s_n = \frac{1}{2} g (2n - 1) \]

### Step 4: Identify the pattern
From the calculations:
- \( s_1 = \frac{g}{2} \)
- \( s_2 = \frac{3g}{2} \)
- \( s_3 = \frac{5g}{2} \)

The distances \( s_1, s_2, s_3, \ldots \) correspond to the odd integers \( 1, 3, 5, \ldots \) multiplied by \( \frac{g}{2} \).

### Step 5: Conclusion
The distances moved by a freely falling body during the 1st, 2nd, 3rd, ..., nth seconds of its motion are proportional to the odd numbers.

### Final Answer
The correct option is that the distances are proportional to odd numbers.

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