To find out when the water balloon hits the ground, we need to determine when the height \(y\) is equal to zero. The equation given is:
\[ y = -6t^2 + 30t + 8 \]
We set \(y\) to zero and solve for \(t\):
\[ 0 = -6t^2 + 30t + 8 \]
Rearranging this, we have:
\[ -6t^2 + 30t + 8 = 0 \]
We can multiply the entire equation by -1 to make the leading coefficient positive:
\[ 6t^2 - 30t - 8 = 0 \]
Next, we can use the quadratic formula to find the values of \(t\), which is given by:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \(a = 6\), \(b = -30\), and \(c = -8\). Now, we can plug these values into the formula:
- Calculate the discriminant \(b^2 - 4ac\):
\[ b^2 - 4ac = (-30)^2 - 4(6)(-8) = 900 + 192 = 1092 \]
- Now plug the values into the quadratic formula:
\[ t = \frac{-(-30) \pm \sqrt{1092}}{2 \cdot 6} \]
This simplifies to:
\[ t = \frac{30 \pm \sqrt{1092}}{12} \]
Now, let's calculate \(\sqrt{1092}\):
\[ \sqrt{1092} \approx 33.0 , (\text{approx}) \]
So now substituting this back into the equation for \(t\):
\[ t = \frac{30 \pm 33.0}{12} \]
This gives us two potential solutions for \(t\):
- \(t = \frac{63}{12} \approx 5.25\)
- \(t = \frac{-3}{12} = -0.25\) (which we discard since time cannot be negative)
Thus, the water balloon will hit the ground approximately at:
\[ t \approx 5.25 \text{ seconds} \]