A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 seconds. It remained in air for :-

Options

A. 16.5 s
B. 17 s
C. 15 s
D. 13 s

Answer

To solve the problem, we need to find the time a stone remains in the air when the distance covered in the last second of its fall is equal to the distance covered in the first 5 seconds.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the total time of fall be \( t \) seconds.
- The distance covered in the last second of motion (i.e., during the \( t \)-th second) is equal to the distance covered in the first 5 seconds.

2. **Distance Covered in the Last Second**:
- The distance covered in the \( t \)-th second can be calculated using the formula:
\[ s_t = u + \frac{1}{2} g (2t - 1) \]
- Since the stone is falling freely from rest, the initial velocity \( u = 0 \):
\[ s_t = \frac{1}{2} g (2t - 1) \]

3. **Distance Covered in the First 5 Seconds**:
- The distance covered in the first 5 seconds can be calculated using the formula:
\[ s_5 = u + \frac{1}{2} g (5^2) \]
- Again, since \( u = 0 \):
\[ s_5 = \frac{1}{2} g (25) = \frac{25g}{2} \]

4. **Setting the Distances Equal**:
- According to the problem, the distance covered in the last second is equal to the distance covered in the first 5 seconds:
\[ \frac{1}{2} g (2t - 1) = \frac{25g}{2} \]

5. **Canceling Common Terms**:
- We can cancel \( \frac{1}{2} g \) from both sides (assuming \( g \neq 0 \)):
\[ 2t - 1 = 25 \]

6. **Solving for \( t \)**:
- Rearranging the equation:
\[ 2t = 25 + 1 = 26 \]
\[ t = \frac{26}{2} = 13 \text{ seconds} \]

### Final Answer:
The stone remains in the air for **13 seconds**.