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. 12s
B. 13s
C. 25s
D. 26s

Explanation

To solve the problem, let's break it down step by step.

### Step 1: Understand the Problem
We need to find the total time \( t \) for which the stone remains in the air. We know that the distance covered in the last second of the motion is equal to the distance covered in the first 5 seconds.

### Step 2: Use the Equation for Distance in Free Fall
The distance \( s \) covered by a freely falling object can be calculated using the formula:
\[ s = ut + \frac{1}{2}gt^2 \]
Since the stone is falling freely from rest, the initial velocity \( u = 0 \). Thus, the formula simplifies to:
\[ s = \frac{1}{2}gt^2 \]

### Step 3: Calculate Distance in the First 5 Seconds
Using the formula for the first 5 seconds:
\[ s_{5} = \frac{1}{2}g(5^2) = \frac{1}{2}g(25) = \frac{25g}{2} \]

### Step 4: Calculate Distance in the Last Second
The distance covered in the last second can be calculated using the formula for distance in the \( n \)-th second:
\[ s_n = u + \frac{1}{2}g(2t - 1) \]
Again, since \( u = 0 \):
\[ s_{last} = \frac{1}{2}g(2t - 1) \]

### Step 5: Set 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:
\[ \frac{1}{2}g(2t - 1) = \frac{25g}{2} \]

### Step 6: Simplify the Equation
We can cancel \( \frac{1}{2}g \) from both sides (assuming \( g \neq 0 \)):
\[ 2t - 1 = 25 \]

### Step 7: Solve for \( t \)
Now, solve for \( t \):
\[ 2t = 25 + 1 \]
\[ 2t = 26 \]
\[ t = \frac{26}{2} = 13 \text{ seconds} \]

### Conclusion
The stone remained in the air for **13 seconds**.

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