Question

y= -6t^2 +30t+8 the water balloon was tossed in the air from the top of a shed
When will the balloon hit the ground

Answer 1

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:

1. Calculate the discriminant \(b^2 - 4ac\):

\[
b^2 - 4ac = (-30)^2 - 4(6)(-8) = 900 + 192 = 1092
\]

2. 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 \(t\):

1. \(t = \frac{63}{12} \approx 5.25\)
2. \(t = \frac{-3}{12} = -0.25\) (which we discard since time cannot be negative)

Thus, the water balloon will hit the ground approximately at:

\[
t \approx 5.25 \text{ seconds}
\]