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 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 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. 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 1

Answer:

h = 8.96 ft.

Explanation:

Given,

*Initial velocity of the ball = v = 26 ft/s

*Distance traveled in front of the foul line = x = 4 ft.

Given curve of the height attained by the basketball is

$\therefore h(x)\ =\ \dfrac{-44x^2}{v^2}\ +\ x\ +\ 6$

substituting the values of  v and x in the above equation of the curve of the height of the ball, we get,

$\Rightarrow h(x)\ =\ \dfrac{-44\times 4^2}{26^2}\ +\ 4\ +\ 6\\\Rightarrow h(x)\ =\ 8.95\ ft.$

Hence the height attained by the ball after the shoot by the basketball player is 8.95 ft up to the ground.