To determine when the water balloon will hit the ground, we need to find when the height \( y \) equals zero. This means we will set the equation \( y = -6t^2 + 30t + 8 \) to zero and solve for \( t \):
\[ -6t^2 + 30t + 8 = 0 \]
This is a standard quadratic equation of the form \( at^2 + bt + c = 0 \), where \( a = -6 \), \( b = 30 \), and \( c = 8 \).
To solve for \( t \), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values:
\[ t = \frac{-30 \pm \sqrt{30^2 - 4 \cdot (-6) \cdot 8}}{2 \cdot (-6)} \] \[ t = \frac{-30 \pm \sqrt{900 + 192}}{-12} \] \[ t = \frac{-30 \pm \sqrt{1092}}{-12} \] \[ t = \frac{-30 \pm 33.0454}{-12} \]
Now we will calculate the two potential solutions:
- \( t = \frac{-30 + 33.0454}{-12} \)
- \( t = \frac{-30 - 33.0454}{-12} \)
Calculating both:
\[ t_1 = \frac{3.0454}{-12} \approx -0.2546 \text{ (not a valid solution)}, \]
\[ t_2 = \frac{-63.0454}{-12} \approx 5.2538. \]
Thus, rounding to the nearest tenth of a second, the balloon will hit the ground at approximately \( t \approx 5.3 \) seconds.
Final Answer:
The balloon will hit the ground at \( 5.3 \) seconds.
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