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
. 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.
Remaining Attempts : 3

Answer 1

To determine when the water balloon hits the ground, we need to find the value of \( t \) when \( y = 0 \) in the given equation:

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

Setting \( y = 0 \):

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

We can rearrange this to:

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

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

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

where \( a = -6 \), \( b = 30 \), and \( c = 8 \).

First, compute the discriminant:

\[ b^2 - 4ac = 30^2 - 4(-6)(8). \]

Calculating \( b^2 \):

\[ 30^2 = 900, \]

Calculating \( 4ac \):

\[ 4 \times -6 \times 8 = -192. \]

So, the discriminant is:

\[ 900 - (-192) = 900 + 192 = 1092. \]

Now, use the quadratic formula:

\[ t = \frac{-30 \pm \sqrt{1092}}{2 \times -6}. \]

Calculating \( \sqrt{1092} \):

\[ \sqrt{1092} \approx 33.0454. \]

Now substitute back into the quadratic formula:

\[ t = \frac{-30 \pm 33.0454}{-12}. \]

This gives us two potential solutions for \( t \):

1. \( t = \frac{-30 + 33.0454}{-12} = \frac{3.0454}{-12} \approx -0.2538 \) (not valid since time cannot be negative).
2. \( t = \frac{-30 - 33.0454}{-12} = \frac{-63.0454}{-12} \approx 5.2538. \)

Rounding \( t = 5.2538 \) to the nearest tenth of a second, we get:

**The balloon will hit the ground at approximately 5.3 seconds.**