Question

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 function, ( h(x)=-\frac{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 with which the ball is shot in feet per second. suppose a player shoots a ball with an initial velocity of 27 feet per second. answer parts (a)-(d).
(a) find the height of the ball after it has traveled 4 feet in front of the foul line.
the height of the ball is 9.03 ft. (round to two decimal places as needed.)
(b) find the height of the ball after it has traveled 9 feet in front of the foul line.
the height of the ball is 10.11 ft. (round to two decimal places as needed.)
(c) find additional points and graph the path of the basketball.

Explanation:

Step1: Substitute \(x = 4\) into the function \(h(x)=-\frac{4x^{2}}{v^{2}}+x + 6\)

Given \(v = 27\), then \(h(4)=-\frac{4\times4^{2}}{27^{2}}+4 + 6\)
First, calculate \(\frac{4\times4^{2}}{27^{2}}=\frac{4\times16}{729}=\frac{64}{729}\approx0.088\)
So \(h(4)=- 0.088+4 + 6\)

Step2: Calculate the value of \(h(4)\)

\(h(4)=-0.088 + 10=9.912\approx9.91\)

Step3: Substitute \(x = 9\) into the function \(h(x)=-\frac{4x^{2}}{v^{2}}+x + 6\)

\(h(9)=-\frac{4\times9^{2}}{27^{2}}+9 + 6\)
Calculate \(\frac{4\times9^{2}}{27^{2}}=\frac{4\times81}{729}=\frac{324}{729}=\frac{4}{9}\approx0.444\)
So \(h(9)=-0.444+9 + 6\)

Step4: Calculate the value of \(h(9)\)

\(h(9)=-0.444 + 15=14.556\approx14.56\)

Step5: Analyze the graph

The function \(h(x)=-\frac{4x^{2}}{v^{2}}+x + 6\) is a quadratic function. The general form of a quadratic function is \(y = ax^{2}+bx + c\), here \(a=-\frac{4}{v^{2}}<0\), so the graph is a parabola opening downwards.

Answer:

(a) \(9.91\) ft
(b) \(14.56\) ft
(c) Graph A (since the function \(h(x)\) is a parabola opening downwards)