Math Problem Statement

PROBLEM

Mark and James love to play volleyball. Suppose Mark tossed a volleyball to James who is 12 ft away. Midway between them is a volleyball net, the top of which is 7 ft high. The ceiling of the gymnasium is 25 ft high. When Mark tossed the ball to James, the shape of its path forms a parabola. The volleyball is 4 ft off the floor when it leaves Mark's hand, and James catches the volleyball 4 ft off the floor as well.

Write the equation that models the path of the volleyball if it passes midway between the net and the celling. Assume the origin is on the point midway through the volleyball court.

Solution

To model the path of the volleyball as a parabolic equation, we need to determine the equation of the parabola that fits the given conditions. Here's how we can proceed:

Problem Recap:

  • The volleyball is tossed by Mark and caught by James, who are 12 feet apart.
  • The net is located midway between them, i.e., 6 feet from both Mark and James.
  • The height of the net is 7 feet.
  • The volleyball is released and caught at a height of 4 feet.
  • The ceiling height is 25 feet, and the ball passes midway between the net and the ceiling (i.e., at 16 feet).

Coordinate System:

We'll place the origin of the coordinate system at the midpoint of the court:

  • The horizontal distance is xx, with x=0x = 0 at the net (midpoint).
  • Mark is at x=6x = -6 (left of the net), and James is at x=6x = 6 (right of the net).
  • The vertical height is yy, representing the height of the ball above the floor.

Key Points:

  1. Starting point (Mark's hand): (6,4)(-6, 4)
  2. Ending point (James' hand): (6,4)(6, 4)
  3. Vertex (top of the parabola): The ball passes at its maximum height midway between the net and the ceiling. This point is (0,16)(0, 16).

Parabolic Equation:

The general form of a parabolic equation is: y=ax2+bx+cy = ax^2 + bx + c

Since the vertex is at (0,16)(0, 16), we can rewrite the equation in vertex form: y=a(xh)2+ky = a(x - h)^2 + k where (h,k)(h, k) is the vertex of the parabola. Here, h=0h = 0 and k=16k = 16. Thus, the equation becomes: y=ax2+16y = ax^2 + 16

Using Points to Find aa:

We substitute the coordinates of either Mark's or James' position to find aa.

Using (6,4)(-6, 4):

4=a(6)2+164 = a(-6)^2 + 16 4=36a+164 = 36a + 16 36a=41636a = 4 - 16 36a=1236a = -12 a=13a = -\frac{1}{3}

Final Equation:

Thus, the equation that models the path of the volleyball is: y=13x2+16\boxed{y = -\frac{1}{3}x^2 + 16}

This equation represents the path of the volleyball in the given scenario.

Would you like more details, or do you have any questions?


Here are 5 related questions you might find interesting:

  1. How would the equation change if the ball were caught at a different height?
  2. What would the equation be if the volleyball was thrown higher initially?
  3. How can we find the time it takes for the ball to travel from Mark to James?
  4. How does the distance between the players affect the shape of the parabola?
  5. How would the equation change if the origin was placed at Mark's location?

Tip: The vertex form of a parabola is particularly useful when you know the highest or lowest point (vertex) of the parabola.

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Math Problem Analysis

Mathematical Concepts

Algebra
Parabolas
Quadratic Equations
Coordinate Geometry

Formulas

Quadratic equation in vertex form: y = a(x - h)^2 + k
Conversion of vertex form to standard form: y = ax^2 + bx + c

Theorems

Vertex Form of a Parabola
Symmetry of Parabolas

Suitable Grade Level

Grades 9-12