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{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 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 ( square mathrm{ft} ). (round to two decimal places as needed.)

Explanation:

Step1: Substitute values into the function

Given \(h(x)=-\frac{44x^{2}}{v^{2}}+x + 6\), \(x = 4\) and \(v=27\).
Substitute into the formula: \(h(4)=-\frac{44\times4^{2}}{27^{2}}+4 + 6\).
First calculate \(44\times4^{2}=44\times16 = 704\) and \(27^{2}=729\).
So \(h(4)=-\frac{704}{729}+4 + 6\).

Step2: Simplify the expression

\(-\frac{704}{729}\approx - 0.966\).
Then \(h(4)\approx-0.966+4 + 6\).
\(h(4)\approx9.034\).

Answer:

\(9.03\)