Question

Use the quadratic formula to calculate the time it will take for the water balloon to hit the ground. Round to the nearest tenth, if necessary.(1 point)
The water balloon will hit the ground at about
seconds.

Answer 1

The flight of a water balloon is modeled by the function h=−2t2+8t+2 where h is the height of the balloon above the ground after t seconds.

Use the vertex formula to determine how many seconds it will take the water balloon to reach the maximum height. Then find the maximum height of the water balloon.

(2 points)
The water balloon will reach its maximum height at
seconds.

The maximum height of the water balloon will be
feet.

Question 2
Use the quadratic formula to calculate the time it will take for the water balloon to hit the ground. Round to the nearest tenth, if necessary.(1 point)
The water balloon will hit the ground at about
seconds.

Answer 2

To find the time it takes for a water balloon to hit the ground using the quadratic formula, we first need to establish the quadratic equation that describes its position as a function of time.

Assuming a water balloon is thrown from a height, its height \( h \) at time \( t \) can typically be represented by the equation:

\[ h(t) = -at^2 + bt + c \]

where:
- \( a \) is the acceleration due to gravity (approximately \( 4.9 \, \text{m/s}^2 \) when considering the downward direction),
- \( b \) is the initial velocity (in positive direction),
- \( c \) is the initial height (in meters).

To find when the balloon hits the ground, we set the height \( h(t) \) equal to 0:

\[ 0 = -at^2 + bt + c \]

Now, we can apply the quadratic formula:

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

To solve this equation, we need specific values for \( a \), \( b \), and \( c \). If you have these values, please provide them so I can help you find the time it will take for the water balloon to hit the ground.

Answer 3

To solve the problem, let's first analyze the given quadratic function for the water balloon:

\[ h(t) = -2t^2 + 8t + 2 \]

### Question 1: Maximum Height

**Finding Time to Reach Maximum Height:**

The time at which a quadratic equation reaches its maximum (or minimum) can be found using the vertex formula:

\[
t = -\frac{b}{2a}
\]

Here:
- \( a = -2 \)
- \( b = 8 \)

Substituting the values:

\[
t = -\frac{8}{2 \cdot -2} = -\frac{8}{-4} = 2 \, \text{seconds}
\]

**Finding the Maximum Height:**

Now we can find the maximum height by substituting \( t = 2 \) into the original height equation:

\[
h(2) = -2(2)^2 + 8(2) + 2
\]
\[
h(2) = -2(4) + 16 + 2
\]
\[
h(2) = -8 + 16 + 2
\]
\[
h(2) = 10 \, \text{feet}
\]

### Summary of Question 1:

- The water balloon will reach its maximum height at **2 seconds**.
- The maximum height of the water balloon will be **10 feet**.

---

### Question 2: Time to Hit the Ground

To find the time it takes for the water balloon to hit the ground, we set the height \( h(t) \) to 0:

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

Rearranging gives:

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

Dividing the entire equation by -2 simplifies it to:

\[
t^2 - 4t - 1 = 0
\]

Now we can use the quadratic formula:

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

Substituting:
- \( a = 1 \)
- \( b = -4 \)
- \( c = -1 \)

Calculating the discriminant:

\[
b^2 - 4ac = (-4)^2 - 4(1)(-1) = 16 + 4 = 20
\]

Now substituting into the quadratic formula:

\[
t = \frac{-(-4) \pm \sqrt{20}}{2 \cdot 1}
\]
\[
t = \frac{4 \pm \sqrt{20}}{2}
\]
\[
t = \frac{4 \pm 2\sqrt{5}}{2}
\]
\[
t = 2 \pm \sqrt{5}
\]

Calculating \( \sqrt{5} \) approximately as 2.236, we have:

\[
t \approx 2 \pm 2.236
\]

This gives us two options:

1. \( t \approx 2 + 2.236 \approx 4.236 \)
2. \( t \approx 2 - 2.236 \) (which will be negative and not applicable for time)

Thus, we take:

\[
t \approx 4.236
\]

**Rounding to the nearest tenth**:

\( t \approx 4.2 \, \text{seconds} \)

### Summary of Question 2:

- The water balloon will hit the ground at about **4.2 seconds**.