Quadratic equations word problem worksheet with examples involving a penny toss.

Problem Analysis:

The problem involves a quadratic equation that models the height of a penny tossed off a bridge into the water. The equation given is:

\[
y = -16t^2 + 96t + 550
\]

where:
- \( y \) is the height of the penny above the water in feet,
- \( t \) is the time in seconds after the penny is tossed.

We need to solve three questions based on this equation:
1. What is the greatest height of the penny?
2. At what time does the penny reach its maximum height?
3. How much time will it take for the coin to hit the water?

---

Solution:

#### Step 1: Understand the Quadratic Equation
The general form of a quadratic equation is:
\[
y = at^2 + bt + c
\]
For our equation:
\[
y = -16t^2 + 96t + 550
\]
we have:
\[
a = -16, \quad b = 96, \quad c = 550
\]

Since \( a < 0 \), the parabola opens downward, meaning the vertex represents the maximum point (the greatest height).

---

#### Question 1: What is the greatest height of the penny?

The greatest height corresponds to the \( y \)-coordinate of the vertex of the parabola. The formula for the \( t \)-coordinate of the vertex (axis of symmetry) is:
\[
t = -\frac{b}{2a}
\]

Substitute \( a = -16 \) and \( b = 96 \):
\[
t = -\frac{96}{2(-16)} = -\frac{96}{-32} = 3
\]

So, the penny reaches its maximum height at \( t = 3 \) seconds. To find the maximum height, substitute \( t = 3 \) back into the original equation:
\[
y = -16(3)^2 + 96(3) + 550
\]
\[
y = -16(9) + 288 + 550
\]
\[
y = -144 + 288 + 550
\]
\[
y = 694
\]

Thus, the greatest height of the penny is:
\[
\boxed{694}
\]

---

#### Question 2: At what time does the penny reach its maximum height?

From the calculation in Question 1, we already determined that the penny reaches its maximum height at:
\[
t = 3
\]

Thus, the time at which the penny reaches its maximum height is:
\[
\boxed{3}
\]

---

#### Question 3: How much time will it take for the coin to hit the water?

When the penny hits the water, its height \( y \) is 0. We need to solve the equation:
\[
-16t^2 + 96t + 550 = 0
\]

This is a quadratic equation. We can solve it using the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = -16 \), \( b = 96 \), and \( c = 550 \).

First, calculate the discriminant:
\[
b^2 - 4ac = 96^2 - 4(-16)(550)
\]
\[
b^2 - 4ac = 9216 + 35200
\]
\[
b^2 - 4ac = 44416
\]

Now, apply the quadratic formula:
\[
t = \frac{-96 \pm \sqrt{44416}}{2(-16)}
\]
\[
t = \frac{-96 \pm 211}{-32}
\]

This gives two solutions:
\[
t = \frac{-96 + 211}{-32} = \frac{115}{-32} = -3.59375 \quad (\text{not valid, as time cannot be negative})
\]
\[
t = \frac{-96 - 211}{-32} = \frac{-307}{-32} = 9.59375
\]

Rounding to the nearest tenth:
\[
t \approx 9.6
\]

Thus, the time it takes for the penny to hit the water is:
\[
\boxed{9.6}
\]

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Final Answers:

1. The greatest height of the penny is \(\boxed{694}\).
2. The penny reaches its maximum height at \(\boxed{3}\) seconds.
3. The time it takes for the penny to hit the water is \(\boxed{9.6}\) seconds.