Question

Zech tossed a water balloon into the air from the top of his shed in the backyard. Use GeoGebra to graph the situation using the formula y=−6t2+30t+8
𝑦
=
−
6
𝑡
2
+
30
𝑡
+
8
. When will the balloon hit the ground? Round your answer to the nearest tenth of a second.(1 point)
The balloon will hit the ground at seconds.

Answer 1

To determine when the water balloon will hit the ground, we need to find when the height \( y \) equals zero. This means we will set the equation \( y = -6t^2 + 30t + 8 \) to zero and solve for \( t \):

\[
-6t^2 + 30t + 8 = 0
\]

This is a standard quadratic equation of the form \( at^2 + bt + c = 0 \), where \( a = -6 \), \( b = 30 \), and \( c = 8 \).

To solve for \( t \), we can use the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Plugging in the values:

\[
t = \frac{-30 \pm \sqrt{30^2 - 4 \cdot (-6) \cdot 8}}{2 \cdot (-6)}
\]
\[
t = \frac{-30 \pm \sqrt{900 + 192}}{-12}
\]
\[
t = \frac{-30 \pm \sqrt{1092}}{-12}
\]
\[
t = \frac{-30 \pm 33.0454}{-12}
\]

Now we will calculate the two potential solutions:

1. \( t = \frac{-30 + 33.0454}{-12} \)
2. \( t = \frac{-30 - 33.0454}{-12} \)

Calculating both:

1.

\[
t_1 = \frac{3.0454}{-12} \approx -0.2546 \text{ (not a valid solution)},
\]

2.

\[
t_2 = \frac{-63.0454}{-12} \approx 5.2538.
\]

Thus, rounding to the nearest tenth of a second, the balloon will hit the ground at approximately \( t \approx 5.3 \) seconds.

**Final Answer:**

The balloon will hit the ground at \( 5.3 \) seconds.