Question

1. nathan kicks a soccer ball and the ball follows a parabolic path. it is in the air for a total of 4 seconds and reaches a maximum height of 20 ft.

a) make a graph of this trajectory:
(graph with y-axis from 0 to 24 and x-axis from 0 to 6)
b) make an equation to model this:
c) use desmos to determine the length of time the ball is above 15 ft.

Explanation:

Part (a)

Step1: Identify Key Points

The ball is in the air for 4 seconds, so it starts at \( (0, 0) \) (time 0, height 0) and lands at \( (4, 0) \) (time 4, height 0). The maximum height is 20 ft at the midpoint of the time interval. The midpoint of 0 and 4 is \( t = 2 \), so the vertex is \( (2, 20) \).

Step2: Plot Points and Draw Parabola

Plot the points \( (0, 0) \), \( (2, 20) \), and \( (4, 0) \). Then draw a parabola (a U - shaped curve) passing through these points. The parabola should open downward since the ball reaches a maximum height and then falls back down.

Part (b)

Step1: Choose Vertex Form of Parabola

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \( (h, k) \) is the vertex. Here, \( h = 2 \) and \( k = 20 \), so the equation becomes \( y = a(x - 2)^2 + 20 \).

Step2: Find the Value of \( a \)

We know that the parabola passes through \( (0, 0) \). Substitute \( x = 0 \) and \( y = 0 \) into the equation:

$$
LATEXBLOCK0
$$

Step3: Write the Equation

Substitute \( a=-5 \), \( h = 2 \), and \( k = 20 \) into the vertex form. The equation is \( y=-5(x - 2)^2+20 \). We can also expand it to standard form:

$$
LATEXBLOCK1
$$

Part (c)

Step1: Set Up the Inequality

We want to find the values of \( x \) (time) when \( y>15 \). So we set up the inequality \( - 5(x - 2)^2+20>15 \).

Step2: Solve the Inequality

$$
LATEXBLOCK0
$$

Step3: Find the Length of Time

To find the length of time the ball is above 15 ft, we calculate the difference between the two bounds: \( 3 - 1=2 \) seconds.

Part (a) Answer:

A parabola with vertex at \( (2, 20) \), passing through \( (0, 0) \) and \( (4, 0) \) (graph drawn as described).

Part (b) Answer:

The equation is \( y=-5(x - 2)^2 + 20 \) (or \( y=-5x^{2}+20x \)).

Part (c) Answer:

The ball is above 15 ft for 2 seconds.

Answer:

Step1: Set Up the Inequality

We want to find the values of \( x \) (time) when \( y>15 \). So we set up the inequality \( - 5(x - 2)^2+20>15 \).

Step2: Solve the Inequality

$$
LATEXBLOCK0
$$

Step3: Find the Length of Time

To find the length of time the ball is above 15 ft, we calculate the difference between the two bounds: \( 3 - 1=2 \) seconds.

Part (a) Answer:

A parabola with vertex at \( (2, 20) \), passing through \( (0, 0) \) and \( (4, 0) \) (graph drawn as described).

Part (b) Answer:

The equation is \( y=-5(x - 2)^2 + 20 \) (or \( y=-5x^{2}+20x \)).

Part (c) Answer:

The ball is above 15 ft for 2 seconds.