A body is dropped from the top of the tower and falls freely.
The distance coverd by it after `n` seconds is directly proportional to .

Options

A. `n^(2)`
B. `n`
C. `2n-1`
D. `2n^(2) -1`

Explanation

To solve the problem, we need to determine how the distance covered by a body falling freely from a height is related to the time elapsed.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The body is dropped from rest, which means its initial velocity \( u = 0 \, \text{m/s} \).
- The only acceleration acting on the body is due to gravity, denoted as \( g \).

2. **Using the Equation of Motion**:
- We can use the second equation of motion, which states:
\[ s = ut + \frac{1}{2} a t^2 \]
- Here, \( s \) is the distance covered, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.

3. **Substituting the Known Values**:
- Since \( u = 0 \) (the body is dropped), the equation simplifies to:
\[ s = 0 \cdot t + \frac{1}{2} g t^2 \]
- This further simplifies to:
\[ s = \frac{1}{2} g t^2 \]

4. **Expressing Distance in Terms of Time**:
- If we let \( t = n \) seconds, we can rewrite the equation as:
\[ s = \frac{1}{2} g n^2 \]

5. **Identifying the Proportionality**:
- From the equation \( s = \frac{1}{2} g n^2 \), we can see that the distance \( s \) is directly proportional to \( n^2 \):
\[ s \propto n^2 \]

6. **Conclusion**:
- Therefore, the distance covered by the body after \( n \) seconds is directly proportional to \( n^2 \).

### Final Answer:
The distance covered by the body after \( n \) seconds is directly proportional to \( n^2 \).