Math Problem Statement

A soccer ball is kicked from the ground. After travelling a horizontal distance of 35 m, it just passes over a 1.5 m tall fence before hitting the ground 38 m from where it was kicked. a) Considering the ground to be the X axis and the vertex to be on the Y axis, determine the equation of a quadratic function that can be used to model the parabolic path of the ball. b) Determine the maximum height of the ball. c) How far has the ball travelled horizontally to reach the maximum height?

Solution

The problem involves modeling the parabolic path of a soccer ball. We can use a quadratic function to describe this motion, as the ball follows a projectile trajectory. Let's break down the solution step by step.

a) Determine the equation of the quadratic function that models the path of the ball

We are given that:

  • The ball is kicked from the ground, so the ball passes through the point (0,0)(0, 0) (this is one root of the quadratic function).
  • The ball hits the ground at (38,0)(38, 0) (this is the other root).
  • The ball passes over a 1.5 m tall fence when it has travelled 35 m horizontally, meaning the point (35,1.5)(35, 1.5) lies on the parabola.

The general form of a quadratic function is: y=ax(x38)y = ax(x - 38) where aa is a constant to be determined, and x=0x = 0 and x=38x = 38 are the roots.

Now, substitute (35,1.5)(35, 1.5) into the equation to find aa:

1.5=a(35)(3538)1.5 = a(35)(35 - 38) 1.5=a(35)(3)1.5 = a(35)(-3) 1.5=105a1.5 = -105a a=1.5105=170a = -\frac{1.5}{105} = -\frac{1}{70}

Thus, the equation of the parabola is: y=170x(x38)y = -\frac{1}{70}x(x - 38)

b) Determine the maximum height of the ball

The maximum height occurs at the vertex of the parabola. For a parabola of the form y=a(xr1)(xr2)y = a(x - r_1)(x - r_2), the vertex is located at the midpoint between the roots r1=0r_1 = 0 and r2=38r_2 = 38.

The horizontal position of the vertex is: xvertex=0+382=19 metersx_{\text{vertex}} = \frac{0 + 38}{2} = 19 \text{ meters}

Substitute x=19x = 19 into the equation to find the maximum height: y=170(19)(1938)y = -\frac{1}{70}(19)(19 - 38) y=170(19)(19)y = -\frac{1}{70}(19)(-19) y=36170y = \frac{361}{70} y5.157 metersy \approx 5.157 \text{ meters}

So, the maximum height of the ball is approximately 5.16 meters.

c) How far has the ball travelled horizontally to reach the maximum height?

As calculated earlier, the horizontal distance to the vertex is: x=19 metersx = 19 \text{ meters}

This is the distance the ball has travelled horizontally to reach its maximum height.

Final answers:

  • a) The equation of the parabolic path is y=170x(x38)y = -\frac{1}{70}x(x - 38).
  • b) The maximum height of the ball is approximately 5.16 meters.
  • c) The ball reaches its maximum height after travelling 19 meters horizontally.

Would you like to explore more details, or have any questions?

Here are 5 follow-up questions that build on this topic:

  1. How would the equation change if the ball passed over the fence at a different height?
  2. What effect would air resistance have on the path of the ball?
  3. How would you calculate the time the ball stays in the air?
  4. How does the initial velocity of the ball relate to the shape of the parabola?
  5. How could you adjust the equation to account for the ball landing at a different point?

Tip: For quadratic functions, the vertex is always halfway between the roots. This can help you quickly find the point of maximum or minimum value.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Projectile Motion

Formulas

Quadratic function: y = ax(x - b)
Midpoint formula: x_vertex = (r1 + r2) / 2

Theorems

Properties of Parabolas

Suitable Grade Level

Grades 10-12