A stone is dropped from a certain heitht which can reach the ground in `5 s`. It is stopped aftre `3 s` of its fall and then it is again released. The total time taken by the stone to reach the ground will be .

Options

A. ` 6s`
B. ` 6.5 s`
C. ` 7 s`
D. ` 7.5 s`

Explanation

To solve the problem step by step, we will analyze the motion of the stone and calculate the total time taken for it to reach the ground after being dropped and then stopped.

### Step-by-Step Solution:

1. **Determine the total height (h) from which the stone is dropped:**
The stone can reach the ground in 5 seconds when dropped from rest. We can use the equation of motion:
\[ h = \frac{1}{2} g t^2 \]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity) and \( t = 5 \, \text{s} \).

Substituting the values:
\[ h = \frac{1}{2} \times 10 \times (5)^2 = \frac{1}{2} \times 10 \times 25 = 125 \, \text{m} \]

2. **Calculate the distance fallen in the first 3 seconds (s1):**
Using the same equation of motion for the first 3 seconds:
\[ s_1 = \frac{1}{2} g t_1^2 \]
where \( t_1 = 3 \, \text{s} \).

Substituting the values:
\[ s_1 = \frac{1}{2} \times 10 \times (3)^2 = \frac{1}{2} \times 10 \times 9 = 45 \, \text{m} \]

3. **Calculate the remaining height to fall (s2):**
The remaining height \( s_2 \) after 3 seconds is:
\[ s_2 = h - s_1 = 125 \, \text{m} - 45 \, \text{m} = 80 \, \text{m} \]

4. **Calculate the time taken to fall the remaining height (t2):**
We can use the equation of motion again to find the time taken to fall the remaining height \( s_2 \):
\[ t_2 = \sqrt{\frac{2 s_2}{g}} \]
Substituting the values:
\[ t_2 = \sqrt{\frac{2 \times 80}{10}} = \sqrt{16} = 4 \, \text{s} \]

5. **Calculate the total time taken (T):**
The total time taken by the stone to reach the ground is the sum of the time taken to fall the first 3 seconds and the time taken to fall the remaining height:
\[ T = t_1 + t_2 = 3 \, \text{s} + 4 \, \text{s} = 7 \, \text{s} \]

### Final Answer:
The total time taken by the stone to reach the ground is **7 seconds**.