To solve the problem, we need to determine the time taken by the stone to reach the ground after being dropped from a height of 39.2 m. The balloon rises with a constant acceleration of 1 m/s².
### Step-by-Step Solution:
1. **Calculate the velocity of the balloon at 39.2 m height:**
We can use the third equation of motion:
\[
v^2 = u^2 + 2as
\]
Here, \(u = 0\) (initial velocity), \(a = 1 \, \text{m/s}^2\) (acceleration), and \(s = 39.2 \, \text{m}\) (height).
\[
v^2 = 0 + 2 \times 1 \times 39.2
\]
\[
v^2 = 78.4
\]
\[
v = \sqrt{78.4} \approx 8.86 \, \text{m/s}
\]
2. **Set up the equation for the stone's motion after being dropped:**
When the stone is dropped, it has an initial velocity \(v\) (upward) and is at a height of 39.2 m. The equation of motion can be written as:
\[
h = vt - \frac{1}{2}gt^2
\]
Here, \(h = 39.2 \, \text{m}\), \(g = 9.8 \, \text{m/s}^2\), and \(v \approx 8.86 \, \text{m/s}\).
3. **Rearranging the equation:**
\[
39.2 = 8.86t - \frac{1}{2} \cdot 9.8t^2
\]
Rearranging gives:
\[
4.9t^2 - 8.86t + 39.2 = 0
\]
4. **Using the quadratic formula to solve for \(t\):**
The quadratic formula is given by:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 4.9\), \(b = -8.86\), and \(c = 39.2\).
\[
t = \frac{-(-8.86) \pm \sqrt{(-8.86)^2 - 4 \cdot 4.9 \cdot 39.2}}{2 \cdot 4.9}
\]
\[
t = \frac{8.86 \pm \sqrt{78.4 - 768.16}}{9.8}
\]
\[
t = \frac{8.86 \pm \sqrt{-689.76}}{9.8}
\]
Since the discriminant is negative, we must have made a mistake in the setup. Let's correct it.
5. **Correcting the equation:**
The correct equation should be:
\[
0 = 4.9t^2 - 8.86t - 39.2
\]
Now, applying the quadratic formula:
\[
t = \frac{8.86 \pm \sqrt{(-8.86)^2 - 4 \cdot 4.9 \cdot (-39.2)}}{2 \cdot 4.9}
\]
\[
t = \frac{8.86 \pm \sqrt{78.4 + 768.16}}{9.8}
\]
\[
t = \frac{8.86 \pm \sqrt{846.56}}{9.8}
\]
\[
t = \frac{8.86 \pm 29.14}{9.8}
\]
6. **Calculating the two possible values for \(t\):**
- Positive root:
\[
t = \frac{38}{9.8} \approx 3.88 \, \text{s}
\]
- Negative root (not physically meaningful):
\[
t = \frac{-20.28}{9.8} \text{ (discarded)}
\]
7. **Final answer:**
The time taken by the stone to reach the ground is approximately \(3.88 \, \text{s}\).
