If a basketball player shoots a foul​ shot, releasing the ball at a​ 45-degree angle from a position 6 feet above the​ floor, then the path of the ball can be modeled by the quadratic​ function,

h left parenthesis x right parenthesis equals negative StartFraction 44 x squared Over v squared EndFraction plus x plus 6 commah(x)=−44x2v2+x+6,

where h is the height of the ball above the​ floor, x is the forward distance of the ball in front of the foul​ line, and v is the initial velocity with which the ball is shot in feet per second.  Suppose a player shoots a ball with an initial velocity of

2727

feet per second.  Answer parts​ (a)-(e) below.

To solve this problem, we'll analyze the quadratic function provided, using the given initial velocity to answer each part. The quadratic function modeling the path of the ball is:

\[
h(x) = -\frac{44x^2}{v^2} + x + 6
\]

where:
- \( h(x) \) is the height of the ball above the floor,
- \( x \) is the forward distance of the ball in front of the foul line,
- \( v \) is the initial velocity in feet per second.

Given:
- Initial velocity \( v = 27 \) feet per second.

### Step-by-Step Analysis

#### (a) **Substitute the initial velocity and simplify the equation.**

Given \( v = 27 \), substitute it into the equation:

\[
h(x) = -\frac{44x^2}{27^2} + x + 6
\]

Simplify:

\[
h(x) = -\frac{44x^2}{729} + x + 6
\]

#### (b) **Determine the maximum height of the ball and the distance from the foul line where this occurs.**

The maximum height occurs at the vertex of the parabola. For a quadratic equation of the form \( h(x) = ax^2 + bx + c \), the x-coordinate of the vertex is given by:

\[
x = -\frac{b}{2a}
\]

In our equation:
- \( a = -\frac{44}{729} \)
- \( b = 1 \)

Substituting:

\[
x = -\frac{1}{2 \times -\frac{44}{729}} = \frac{729}{88}
\]

The maximum height can be found by substituting \( x = \frac{729}{88} \) back into the equation for \( h(x) \):

\[
h\left(\frac{729}{88}\right) = -\frac{44 \left(\frac{729}{88}\right)^2}{729} + \frac{729}{88} + 6
\]

#### (c) **Find the height of the ball when it reaches a forward distance of 15 feet.**

Substitute \( x = 15 \) into the equation:

\[
h(15) = -\frac{44(15)^2}{729} + 15 + 6
\]

Calculate the value:

\[
h(15) = -\frac{44 \times 225}{729} + 15 + 6 = -\frac{9900}{729} + 21
\]

#### (d) **Find the distance from the foul line where the ball will hit the floor (i.e., when \( h(x) = 0 \)).**

Set \( h(x) = 0 \) and solve for \( x \):

\[
0 = -\frac{44x^2}{729} + x + 6
\]

Multiply through by 729 to eliminate the fraction:

\[
0 = -44x^2 + 729x + 4374
\]

Use the quadratic formula to solve for \( x \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where:
- \( a = -44 \)
- \( b = 729 \)
- \( c = 4374 \)

#### (e) **Interpret the results in the context of the problem.**

The results should tell us:
- The maximum height of the ball and where it occurs.
- The height at a specific forward distance.
- The point where the ball will land on the floor, representing how far from the foul line it reaches the ground.

Would you like detailed calculations for any of these parts?

If a basketball player shoots a foul shot, releasing the ball at a 45-degree angle from a position 6 feet above the floor, then the path of the ball can be modeled by the quadratic function h(x) = -44x^2/v^2 + x + 6, where h is the height of the ball above the floor, x is the forward distance of the ball in front of the foul line, and v is the initial velocity in feet per second. Suppose a player shoots a ball with an initial velocity of 27 feet per second. Answer parts (a)-(e) below.

Explore how to model a basketball shot with a quadratic function. This example walks through solving for maximum height, forward distance, and projectile motion using the quadratic equation. Suitable for Grades 10-12 students.

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