To find out when the water balloon will hit the ground, we need to determine when the height \( y \) of the balloon equals zero. This is represented by the equation:
\[ y = -6t^2 + 30t + 8 \]
To find when the balloon hits the ground, set \( y = 0 \):
\[ -6t^2 + 30t + 8 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = -6 \), \( b = 30 \), and \( c = 8 \).
First, calculate the discriminant:
\[ b^2 - 4ac = 30^2 - 4 \cdot (-6) \cdot 8 \] \[ = 900 + 192 \] \[ = 1092 \]
Now, plug the values into the quadratic formula:
\[ t = \frac{-30 \pm \sqrt{1092}}{2 \cdot (-6)} \] \[ = \frac{-30 \pm \sqrt{1092}}{-12} \]
Next, calculate \( \sqrt{1092} \):
\[ \sqrt{1092} \approx 33.0 \]
Now substituting back, we get two values for \( t \):
\[ t = \frac{-30 + 33.0}{-12} \quad \text{and} \quad t = \frac{-30 - 33.0}{-12} \]
Calculating these:
- For the positive root:
\[ t_1 = \frac{3.0}{-12} \approx -0.25 \quad \text{(not meaningful in this context)} \]
- For the negative root:
\[ t_2 = \frac{-63.0}{-12} \approx 5.25 \]
Thus, rounding to the nearest tenth of a second, we find:
\[ t \approx 5.3 \text{ seconds} \]
Therefore, the balloon will hit the ground at approximately 5.3 seconds.