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, $h(x)=-\frac{44 x^{2}}{v^{2}}+x+6$, 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 26 feet per second. Answer parts (a)-(e) below.
(a) Determine the height of the ball after it has traveled 4 feet in front of the foul line.
The height of the ball is ft. (Round to two decimal places as needed.)
Answer
Rounding to two decimal places, the height of the ball after it has traveled 4 feet in front of the foul line is approximately \(\boxed{8.96}\) feet.
Step 1 ：Given that the basketball player releases the ball at a 45-degree angle from a position 6 feet above the floor, the path of the ball can be modeled by the quadratic function, \(h(x)=-\frac{44 x^{2}}{v^{2}}+x+6\), 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.
Step 2 ：Suppose a player shoots a ball with an initial velocity of 26 feet per second.
Step 3 ：To determine the height of the ball after it has traveled 4 feet in front of the foul line, we substitute \(x=4\) and \(v=26\) into the equation.
Step 4 ：Substituting these values into the equation gives us \(h = -\frac{44 \times 4^{2}}{26^{2}}+4+6\).
Step 5 ：Solving this equation gives us \(h = 8.958579881656805\).
Step 6 ：Rounding to two decimal places, the height of the ball after it has traveled 4 feet in front of the foul line is approximately \(\boxed{8.96}\) feet.
