A balloon is going upwards with a constant velocity `15 m//s`. When the ballooon is at 50 m height, a stone is dropped outside from the balloon. How long will stone take to reach at the ground? `("take " g=10 m//s^(2))`

Explanation

To solve the problem, we need to determine how long it takes for the stone to reach the ground after being dropped from a balloon that is moving upwards with a constant velocity of 15 m/s from a height of 50 m.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Initial height of the stone, \( h = 50 \, \text{m} \)
- Initial velocity of the stone, \( u = -15 \, \text{m/s} \) (negative because it is moving downward)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) (acting downwards)

2. **Use the Kinematic Equation:**
We will use the kinematic equation:
\[ s = ut + \frac{1}{2}gt^2 \]
where:
- \( s \) is the displacement (which will be -50 m since it falls to the ground),
- \( u \) is the initial velocity,
- \( g \) is the acceleration due to gravity,
- \( t \) is the time in seconds.

3. **Set Up the Equation:**
Plugging in the values:
\[ -50 = -15t + \frac{1}{2}(10)t^2 \]
Simplifying gives:
\[ -50 = -15t + 5t^2 \]
Rearranging the equation:
\[ 5t^2 - 15t + 50 = 0 \]

4. **Rearranging the Equation:**
To make it easier to solve, we can divide the entire equation by 5:
\[ t^2 - 3t - 10 = 0 \]

5. **Factor the Quadratic Equation:**
We can factor this equation:
\[ (t - 5)(t + 2) = 0 \]

6. **Solve for \( t \):**
Setting each factor to zero gives us:
\[ t - 5 = 0 \quad \Rightarrow \quad t = 5 \, \text{s} \]
\[ t + 2 = 0 \quad \Rightarrow \quad t = -2 \, \text{s} \quad (\text{not valid as time cannot be negative}) \]

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