beck tosses a ball up in the air. she releases it at a height of 6 ft above the ground. two seconds later, she catches it at the same height, 6 ft. what is a quadratic function to model this?

To model the height of the ball over time, we can use a quadratic function of the form:

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

where:
- \(h(t)\) represents the height of the ball at time \(t\),
- \(a\), \(b\), and \(c\) are constants to be determined based on the given information.

### Step 1: Identify key points

- At \(t = 0\) (when the ball is released), the height is \(6\) ft.
- At \(t = 2\) seconds (when Beck catches the ball), the height is again \(6\) ft.

These give us two points: \( (0, 6) \) and \( (2, 6) \).

Also, since the ball is tossed upwards and then caught back at the same height, it follows a parabolic trajectory that reaches a maximum height at the midpoint between these times (i.e., at \(t = 1\) second).

### Step 2: Use known points to determine the constants

We know:
1. \(h(0) = 6\), so substitute \(t = 0\) and \(h(0) = 6\) into the equation:

\[
6 = a(0)^2 + b(0) + c \quad \Rightarrow \quad c = 6
\]

Now the function becomes:

\[
h(t) = at^2 + bt + 6
\]

2. At \(t = 2\), the height is again \(6\):

\[
6 = a(2)^2 + b(2) + 6
\]

Simplifying:

\[
6 = 4a + 2b + 6
\]
\[
0 = 4a + 2b \quad \Rightarrow \quad 2a + b = 0 \quad \Rightarrow \quad b = -2a
\]

3. To find \(a\), we use the fact that the vertex of the parabola occurs at \(t = 1\) (the midpoint between \(t = 0\) and \(t = 2\), where the ball reaches its maximum height). At the vertex, the derivative of the function equals zero. The derivative of the function \(h(t)\) is:

\[
h'(t) = 2at + b
\]

At \(t = 1\), the derivative equals zero:

\[
0 = 2a(1) + b \quad \Rightarrow \quad 2a + b = 0
\]

This is the same equation we derived earlier, confirming that \(b = -2a\).

### Step 3: Solve for \(a\)

We need another piece of information to solve for \(a\). Given that the maximum height occurs at \(t = 1\), we can assume that the ball's height at \(t = 1\) is the vertex of the parabola. However, if the exact maximum height is not given, we can use symmetry of the trajectory, assuming that the parabolic path makes the ball come back to the same height at \(t = 0\) and \(t = 2\) with no further details.

### Final Equation:

Thus, the quadratic function that models the height of the ball is:

\[
h(t) = -2t^2 + 4t + 6
\]

This equation assumes symmetry and models the ball’s height with respect to time as it travels upward and downward.

---

**Would you like more details or a breakdown of the steps?**

Here are some related questions for further exploration:
1. How do you determine the vertex of a quadratic function?
2. What happens if the maximum height of the ball is known?
3. How do we model projectile motion using a quadratic function in general?
4. What are the differences between the coefficients in a quadratic function and their impact on the graph?
5. How do we derive the equation of motion for an object thrown vertically in physics?

**Tip:** If you have the equation of a parabola, the vertex form \(h(t) = a(t - h)^2 + k\) can often make it easier to directly see the maximum or minimum point of the motion!

Beck tosses a ball up in the air. She releases it at a height of 6 ft above the ground. Two seconds later, she catches it at the same height, 6 ft. What is a quadratic function to model this?

Learn how to model the height of a ball tossed in the air using a quadratic function. This problem covers concepts like projectile motion and how to derive a quadratic equation from given time and height data. The equation used is h(t) = -2t^2 + 4t + 6, assuming symmetry and a maximum height at the midpoint.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation