To determine when the water balloon hits the ground, we need to find the value of \( t \) when \( y = 0 \) in the given equation:
\[ y = -6t^2 + 30t + 8. \]
Setting \( y = 0 \):
\[ 0 = -6t^2 + 30t + 8. \]
We can rearrange this to:
\[ -6t^2 + 30t + 8 = 0. \]
To solve for \( t \), we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = -6 \), \( b = 30 \), and \( c = 8 \).
First, compute the discriminant:
\[ b^2 - 4ac = 30^2 - 4(-6)(8). \]
Calculating \( b^2 \):
\[ 30^2 = 900, \]
Calculating \( 4ac \):
\[ 4 \times -6 \times 8 = -192. \]
So, the discriminant is:
\[ 900 - (-192) = 900 + 192 = 1092. \]
Now, use the quadratic formula:
\[ t = \frac{-30 \pm \sqrt{1092}}{2 \times -6}. \]
Calculating \( \sqrt{1092} \):
\[ \sqrt{1092} \approx 33.0454. \]
Now substitute back into the quadratic formula:
\[ t = \frac{-30 \pm 33.0454}{-12}. \]
This gives us two potential solutions for \( t \):
- \( t = \frac{-30 + 33.0454}{-12} = \frac{3.0454}{-12} \approx -0.2538 \) (not valid since time cannot be negative).
- \( t = \frac{-30 - 33.0454}{-12} = \frac{-63.0454}{-12} \approx 5.2538. \)
Rounding \( t = 5.2538 \) to the nearest tenth of a second, we get:
The balloon will hit the ground at approximately 5.3 seconds.
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