A stone falls freely rest. The distance covered by it in the last second is equal to the distance covered by it in the first 2 s. The time taken by the stone to reach the ground is

Options

A. `2.5 s`
B. `3.5 s`
C. `4 s`
D. `5 s`

Explanation

To solve the problem, we will use the equations of motion under uniform acceleration due to gravity. The stone is falling freely from rest, so we can apply the following kinematic equations:

1. The distance covered in time \( t \) is given by:
\[ s = ut + \frac{1}{2}gt^2 \]
where \( u \) is the initial velocity (which is 0), \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)), and \( t \) is the time in seconds.

2. The distance covered in the last second of fall can be calculated using the formula:
\[ s_n = u + \frac{1}{2}g(2n - 1) \]
where \( n \) is the total time of fall in seconds.

3. The distance covered in the first 2 seconds is:
\[ s_2 = u \cdot 2 + \frac{1}{2}g(2^2) = 0 \cdot 2 + \frac{1}{2}g(4) = 2g \]

4. According to the problem, the distance covered in the last second is equal to the distance covered in the first 2 seconds. Thus:
\[ s_n = s_2 \]

5. We can express \( s_n \) as:
\[ s_n = \frac{1}{2}g(n^2) - \frac{1}{2}g((n-1)^2) = \frac{1}{2}g(n^2 - (n^2 - 2n + 1)) = \frac{1}{2}g(2n - 1) \]

6. Setting \( s_n = s_2 \):
\[ \frac{1}{2}g(2n - 1) = 2g \]

7. Dividing both sides by \( g \) and multiplying by 2:
\[ 2n - 1 = 4 \]

8. Solving for \( n \):
\[ 2n = 5 \implies n = \frac{5}{2} = 2.5 \, \text{s} \]

Thus, the time taken by the stone to reach the ground is \( 2.5 \, \text{s} \).