A body is dropped from the top of the tower and falls freely.
The distance coverd in the `nth` second is proportilnal to .

Options

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

Explanation

To solve the problem of finding the distance covered by a body dropped from a tower in the nth second, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body is dropped from rest, meaning its initial velocity (u) is 0.
- We need to find the distance covered in the nth second.

2. **Distance in n Seconds**:
- The formula for the distance covered by a freely falling body in n seconds is given by:
\[ d_n = ut + \frac{1}{2}gt^2 \]
- Since the initial velocity (u) is 0, this simplifies to:
\[ d_n = \frac{1}{2}gn^2 \]

3. **Distance in (n-1) Seconds**:
- Similarly, the distance covered in (n-1) seconds is:
\[ d_{n-1} = u(n-1) + \frac{1}{2}g(n-1)^2 \]
- Again, since u = 0, this simplifies to:
\[ d_{n-1} = \frac{1}{2}g(n-1)^2 \]

4. **Calculating Distance in the nth Second**:
- The distance covered in the nth second (denoted as \( S_n \)) is the difference between the distance covered in n seconds and the distance covered in (n-1) seconds:
\[ S_n = d_n - d_{n-1} \]
- Substituting the expressions we derived:
\[ S_n = \frac{1}{2}gn^2 - \frac{1}{2}g(n-1)^2 \]

5. **Simplifying the Expression**:
- Expanding \( (n-1)^2 \):
\[ (n-1)^2 = n^2 - 2n + 1 \]
- Therefore:
\[ S_n = \frac{1}{2}gn^2 - \frac{1}{2}g(n^2 - 2n + 1) \]
- Simplifying further:
\[ S_n = \frac{1}{2}gn^2 - \frac{1}{2}gn^2 + g(n - \frac{1}{2}) \]
- This simplifies to:
\[ S_n = g\left(n - \frac{1}{2}\right) \]

6. **Final Expression**:
- Thus, the distance covered in the nth second is:
\[ S_n = g\left(2n - 1\right) / 2 \]
- Since g is a constant, we can say that the distance covered in the nth second is proportional to \( 2n - 1 \).

### Conclusion:
The distance covered in the nth second is proportional to \( 2n - 1 \).