A ballon is ascending with a constant velocity of 20 m/s. A particle is dropped from it when its height was 60m above the ground. Find if `(g = 10 m//s^(2))`. The time of flight of particle

Options

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

Answer

To solve the problem of a particle dropped from a balloon ascending at a constant velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial height of the balloon (h) = 60 m
- Initial velocity of the particle (u) = 20 m/s (upward)
- Acceleration due to gravity (g) = 10 m/s² (downward)

2. **Set Up the Equation of Motion:**
We will use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, we will consider downward direction as positive, so:
- \( s = -60 \) m (since the particle is falling down)
- \( u = -20 \) m/s (upward velocity is negative)
- \( a = g = 10 \) m/s² (downward acceleration)

Substituting these values into the equation gives:
\[ -60 = -20t + \frac{1}{2} (10) t^2 \]

3. **Simplify the Equation:**
\[ -60 = -20t + 5t^2 \]
Rearranging the equation:
\[ 5t^2 - 20t + 60 = 0 \]
Dividing the entire equation by 5 to simplify:
\[ t^2 - 4t + 12 = 0 \]

4. **Solve the Quadratic Equation:**
We can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -4 \), and \( c = 12 \).
\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \]
\[ t = \frac{4 \pm \sqrt{16 - 48}}{2} \]
\[ t = \frac{4 \pm \sqrt{-32}}{2} \]
Since the discriminant is negative, this means there are no real solutions for time \( t \).

5. **Conclusion:**
Since the quadratic equation does not yield a valid time, we conclude that the particle does not reach the ground within the given parameters.

### Final Answer:
The time of flight of the particle is not applicable as it does not reach the ground.