Solving a quadratic equation to determine the time a rocket reaches 30 feet and when it lands.

Problem Analysis:

The problem involves a quadratic equation that models the height of a rocket as a function of time. The height \( h(t) \) is given by:

\[
h(t) = -16t^2 + 120t + 80
\]

We are tasked with solving two parts:

1. When will the rocket be 30 feet above the sea?
2. When will the rocket fall into the sea (i.e., when will its height be 0)?

Step-by-Step Solution:

#### Part 1: When will the rocket be 30 feet above the sea?

To find when the rocket is 30 feet above the sea, we set \( h(t) = 30 \):

\[
-16t^2 + 120t + 80 = 30
\]

Subtract 30 from both sides to set the equation to zero:

\[
-16t^2 + 120t + 80 - 30 = 0
\]

Simplify:

\[
-16t^2 + 120t + 50 = 0
\]

This is a quadratic equation in the standard form \( at^2 + bt + c = 0 \), where:
\[
a = -16, \quad b = 120, \quad c = 50
\]

We solve this using the quadratic formula:

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

Substitute \( a = -16 \), \( b = 120 \), and \( c = 50 \):

\[
t = \frac{-120 \pm \sqrt{120^2 - 4(-16)(50)}}{2(-16)}
\]

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

\[
b^2 = 120^2 = 14400
\]
\[
4ac = 4(-16)(50) = -3200
\]
\[
b^2 - 4ac = 14400 - (-3200) = 14400 + 3200 = 17600
\]

Now, substitute back into the quadratic formula:

\[
t = \frac{-120 \pm \sqrt{17600}}{-32}
\]

Simplify \( \sqrt{17600} \):

\[
\sqrt{17600} = \sqrt{16 \times 1100} = 4\sqrt{1100} = 4 \times 10\sqrt{11} = 40\sqrt{11}
\]

So:

\[
t = \frac{-120 \pm 40\sqrt{11}}{-32}
\]

Simplify the fraction:

\[
t = \frac{-120}{-32} \pm \frac{40\sqrt{11}}{-32}
\]
\[
t = \frac{120}{32} \mp \frac{40\sqrt{11}}{32}
\]
\[
t = \frac{15}{4} \mp \frac{5\sqrt{11}}{4}
\]

Thus, the two solutions are:

\[
t = \frac{15 + 5\sqrt{11}}{4} \quad \text{and} \quad t = \frac{15 - 5\sqrt{11}}{4}
\]

These are the times when the rocket is 30 feet above the sea.

#### Part 2: When will the rocket fall into the sea?

To find when the rocket falls into the sea, we set \( h(t) = 0 \):

\[
-16t^2 + 120t + 80 = 0
\]

This is another quadratic equation in the standard form \( at^2 + bt + c = 0 \), where:
\[
a = -16, \quad b = 120, \quad c = 80
\]

We solve this using the quadratic formula:

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

Substitute \( a = -16 \), \( b = 120 \), and \( c = 80 \):

\[
t = \frac{-120 \pm \sqrt{120^2 - 4(-16)(80)}}{2(-16)}
\]

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

\[
b^2 = 120^2 = 14400
\]
\[
4ac = 4(-16)(80) = -5120
\]
\[
b^2 - 4ac = 14400 - (-5120) = 14400 + 5120 = 19520
\]

Now, substitute back into the quadratic formula:

\[
t = \frac{-120 \pm \sqrt{19520}}{-32}
\]

Simplify \( \sqrt{19520} \):

\[
\sqrt{19520} = \sqrt{16 \times 1220} = 4\sqrt{1220} = 4 \times 2\sqrt{305} = 8\sqrt{305}
\]

So:

\[
t = \frac{-120 \pm 8\sqrt{305}}{-32}
\]

Simplify the fraction:

\[
t = \frac{-120}{-32} \pm \frac{8\sqrt{305}}{-32}
\]
\[
t = \frac{120}{32} \mp \frac{8\sqrt{305}}{32}
\]
\[
t = \frac{15}{4} \mp \frac{\sqrt{305}}{4}
\]

Thus, the two solutions are:

\[
t = \frac{15 + \sqrt{305}}{4} \quad \text{and} \quad t = \frac{15 - \sqrt{305}}{4}
\]

Since time cannot be negative, we discard the negative solution. Therefore, the time when the rocket falls into the sea is:

\[
t = \frac{15 + \sqrt{305}}{4}
\]

Final Answers:

1. The rocket will be 30 feet above the sea at:
\[
\boxed{\frac{15 + 5\sqrt{11}}{4} \text{ and } \frac{15 - 5\sqrt{11}}{4}}
\]

2. The rocket will fall into the sea at:
\[
\boxed{\frac{15 + \sqrt{305}}{4}}
\]