In a basketball game, Noli scored 3 points by shooting the ball at a horizontal distance of 7 feet from the 10-feet-high hoop. Noli threw the ball 6 feet above the ground and the ball went twice as high as its initial horizontal distance from the hoop. What was the horizontal distance of the ball from Noli when it reached its maximum height?

The horizontal distance of the ball from Noli when it reached maximum height is approximately 4.1 feet

Step-by-step explanation:

The horizontal distance from the hoop at which Noli threw the ball, y₀ = 7 feet

The height of the hoop = 10 feet

The height from which the ball was thrown = 6 feet

The height to which the ball is thrown = 2 × The initial horizontal distance from the hoop

∴ The height to which the ball is thrown, = 2 × 7 feet = 14 feet

The coordinates of points on the path of the ball are;

(0, 6), (7, 10), (x, 14)

The projectile in vertex form is y = a·(x - h)² + k

Where;

(h, k) = The x, and y-coordinate of the maximum height

The y-coordinate of the maximum height reached = k = 14 feet

The x-coordinate of the maximum height = h = The horizontal distance of the ball from Noli when it reached maximum height

At y = 6, x = 0, therefore, we get;

6 = a·(0 - h)² + 14 = a·h² + 14

6 = a·h² + 14

a = -8/h²

At y = 10, x = 7, we get;

10 = a·(7 - h)² + 14 = (-8/h²)·(7 - h)² + 14

10·h² = -8·(7 - h)² + 14·h²

-4·h² =  -8·(7 - h)²

4·h² - 8·(7 - h)² = 0

-4·h² + 112·h - 392 = 0

h = (-112 ± √(112² - 4 × (-4) × (-392)))/(2 × -4)

h ≈ 4.1, or h ≈ 23.9

We reject the value, h ≈ 23.9 feet given that the distance between the where the Noli threw the ball and the hoop = 7 feet

Therefore, the horizontal distance of the ball from Noli at which the ball reached its maximum height, h ≈ 4.1 feet