Help!
Sarah was standing on a dock next to the ocean. She kicked a soccer ball off the dock to her friend in the water. 4 seconds after she kicked the ball it reached a height of 9 feet above the water. It hit the water 6 seconds later.
Write an equation for the path of the rock in y=a(x-h)^2+k and give the height of the dock that she kicked the ball from. Show your work!

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

Step-by-step 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

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