Answer:
1) The equation for the of the ball in y = a·(x - h)² + k is;
y = -16·(x - 5)² + 25
2) The height of the dock, d = -375 feet below the water level
Explanation:
1) The question is with regards to quadratic function representing projectile motion
The given parameters are;
The height the ball reaches 4 seconds after Sarah kicked the ball = 9 feet
The time the ball hits the water = 6 seconds after reaching the 9 feet height
The form of the quadratic equation representing the motion is given as follows;
y = a·(x - h)² + k = a·x² - 2·a·h·x + a·h² + k
Let 'x' represent the time of motion of the ball, and let 'a', represent the acceleration due to gravity, we have;
The equation for the ball, y = a·(x - h)² + k
Where;
(h, k) = The coordinates of the vertex
h = The horizontal component of the vertex coordinate = 0
Therefore, we have;
When x = 0, y = d
d = -16·(0 - h)² + k = -16·h² + k
d = -16·h² + k
When x = 4, y = 9 - d
9 - d = -16(4 - h)² + k = -16(4 - h)² + k
When x = 2, y = d
d = -16(2 - h)² + k
When x = 6, y = 9
9 = -16(6 - 5)² + k
When x = 8, y = d
d = -16(8 - h)² + k
-16(8 - h)² - (-16(2 - h)²) = 0
h = 5
From 9 - d = -16(4 - h)² + k = -16(4 - 4)² + k
d = 9 - k
9 = -16(6 - h)² + k
k = 9 + 16(6 - 5)² = 25
d = 9 - k = 9 - 25 = -16
Therefore, h = 5, k = 25
The equation for the of the ball in y = a·(x - h)² + k is therefore;
y = -16·(x - 5)² + 25
2) When x = 0, y = d, ∴ d = -16(0 - 5)² + 25 = -375 feet below the water
The height of the dock, d = -375 feet below the water level