A ball is dropped from a ballon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take in reaching the ground?

Explanation

To solve the problem of how long it takes for a ball dropped from a balloon (which is moving upwards at a speed of 7 m/s) to reach the ground from a height of 60 m, we can use the equations of motion.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Initial height of the balloon (h) = 60 m
- Initial velocity of the ball (u) = 7 m/s (upwards)
- Acceleration due to gravity (g) = 9.8 m/s² (downwards, but we will take it as -10 m/s² for simplicity)
- Final position (s) = 0 m (ground level)

2. **Set Up the Equation of Motion:**
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, we consider downward direction as negative, so:
\[ -60 = 7t - \frac{1}{2} \cdot 10 \cdot t^2 \]

3. **Rearranging the Equation:**
Rearranging the equation gives:
\[ -60 = 7t - 5t^2 \]
Rearranging further:
\[ 5t^2 - 7t - 60 = 0 \]

4. **Using the Quadratic Formula:**
The quadratic equation is in the form \( at^2 + bt + c = 0 \), where:
- \( a = 5 \)
- \( b = -7 \)
- \( c = -60 \)

We can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

5. **Calculating the Discriminant:**
First, calculate the discriminant:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 5 \cdot (-60) = 49 + 1200 = 1249 \]

6. **Finding the Roots:**
Now substituting into the quadratic formula:
\[ t = \frac{7 \pm \sqrt{1249}}{10} \]
Calculate \( \sqrt{1249} \):
\[ \sqrt{1249} \approx 35.33 \]
So:
\[ t = \frac{7 \pm 35.33}{10} \]

7. **Calculating the Two Possible Values for t:**
- \( t_1 = \frac{7 + 35.33}{10} = \frac{42.33}{10} \approx 4.23 \) seconds
- \( t_2 = \frac{7 - 35.33}{10} = \frac{-28.33}{10} \) (not valid since time cannot be negative)

8. **Final Answer:**
The time taken for the ball to reach the ground is approximately **4.23 seconds**.