To determine when the water balloon will hit the ground, we need to find the time \( t \) when the height \( y \) is equal to zero. The equation you've provided is:
\[ y = -6t^2 + 30t + 8 \]
To find when the balloon hits the ground, we set \( y \) to 0:
\[ 0 = -6t^2 + 30t + 8 \]
We can rearrange this to:
\[ -6t^2 + 30t + 8 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = -6 \), \( b = 30 \), and \( c = 8 \).
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 30^2 - 4(-6)(8) \] \[ = 900 + 192 = 1092 \]
- Now, substitute the values into the quadratic formula:
\[ t = \frac{-30 \pm \sqrt{1092}}{2(-6)} \]
- Calculate \( \sqrt{1092} \):
\[ \sqrt{1092} \approx 33.0 \]
- Now, substitute this back into the formula:
\[ t = \frac{-30 \pm 33.0}{-12} \]
This results in two possible solutions for \( t \):
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First solution (using the positive sign): \[ t = \frac{-30 + 33.0}{-12} = \frac{3.0}{-12} = -0.25 \quad (\text{not a valid solution since } t \text{ cannot be negative}) \]
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Second solution (using the negative sign): \[ t = \frac{-30 - 33.0}{-12} = \frac{-63.0}{-12} = 5.25 \]
Thus, the water balloon will hit the ground approximately \( 5.25 \) seconds after being tossed.
Finally, rounding to the nearest tenth: \[ t \approx 5.3 \text{ seconds} \]
So, the water balloon will hit the ground at approximately 5.3 seconds.
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