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