To determine when the water balloon will hit the ground using the given equation \( y = -6t^2 + 30t + 8 \), we need to find the value of \( t \) when \( y = 0 \).
Setting the equation to zero: \[ -6t^2 + 30t + 8 = 0 \]
We can solve this quadratic equation using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, the coefficients are:
- \( a = -6 \)
- \( b = 30 \)
- \( c = 8 \)
Plugging in these values into the quadratic formula gives: \[ t = \frac{-30 \pm \sqrt{30^2 - 4(-6)(8)}}{2(-6)} \] \[ t = \frac{-30 \pm \sqrt{900 + 192}}{-12} \] \[ t = \frac{-30 \pm \sqrt{1092}}{-12} \]
Calculating \( \sqrt{1092} \): \[ \sqrt{1092} \approx 33.0 \]
So, continuing with our calculation: \[ t = \frac{-30 \pm 33.0}{-12} \]
Now calculating the two possible values for \( t \):
- \( t = \frac{-30 + 33.0}{-12} = \frac{3.0}{-12} = -0.25 \) (not valid, as time cannot be negative)
- \( t = \frac{-30 - 33.0}{-12} = \frac{-63.0}{-12} = 5.25 \)
Thus, rounding to the nearest tenth: \[ t \approx 5.3 \text{ seconds} \]
Therefore, the balloon will hit the ground at approximately 5.3 seconds.
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