A balloon is going vertically upwards with a velocity of `10 m//s`. When it is `75 m` above the ground, a stone is gently relesed from it. The time taken by the stone to reach the ground is `(g=10 m//s^(2))`

Options

A. 3 s
B. 4 s
C. 5 s
D. 6 s

Explanation

To solve the problem, we need to determine the time taken by the stone to reach the ground after being released from the balloon. The balloon is moving upwards with an initial velocity of \(10 \, \text{m/s}\) and is at a height of \(75 \, \text{m}\) above the ground when the stone is released. We will use the equations of motion to find the time taken.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Initial velocity of the stone, \( u = 10 \, \text{m/s} \) (upward).
- Initial height of the stone, \( h = 75 \, \text{m} \) (above the ground).
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (downward, hence we will take it as negative in our calculations).

2. **Set Up the Equation of Motion:**
We will use the second equation of motion:
\[ h = ut + \frac{1}{2} a t^2 \]
Here, \( h \) will be taken as \(-75 \, \text{m}\) (since it is moving down to the ground), \( u = 10 \, \text{m/s} \), and \( a = -g = -10 \, \text{m/s}^2 \).

3. **Substitute the Values:**
\[ -75 = 10t - \frac{1}{2} \cdot 10 \cdot t^2 \]
Simplifying this gives:
\[ -75 = 10t - 5t^2 \]

4. **Rearranging the Equation:**
Rearranging the equation to standard quadratic form:
\[ 5t^2 - 10t - 75 = 0 \]
Dividing the entire equation by 5:
\[ t^2 - 2t - 15 = 0 \]

5. **Factoring the Quadratic Equation:**
We can factor this equation:
\[ (t - 5)(t + 3) = 0 \]

6. **Finding the Roots:**
Setting each factor to zero gives us:
\[ t - 5 = 0 \quad \Rightarrow \quad t = 5 \, \text{s} \]
\[ t + 3 = 0 \quad \Rightarrow \quad t = -3 \, \text{s} \quad (\text{not valid since time cannot be negative}) \]

7. **Conclusion:**
The time taken by the stone to reach the ground is:
\[ t = 5 \, \text{s} \]

### Final Answer:
The time taken by the stone to reach the ground is **5 seconds**.