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Refer to the diagram shown below.
The path of the rock is
y = -0.005x² + 0.4x - 5.1
When x = 0, y = -5.1 ft, which is the location of the hole level.
The rock reaches ground level two times when x = x₁ and when x = x₂.
At ground level, y = 0. Therefore
-0.005x² + 0.4x - 5.1 = 0
Divide through by -0.005.
x² - 82x + 1020 = 0
Solve with the quadratic formula.
x = (82 +/- √(82² - 4*1020))/2
= (82 +/- 51.4198)/2
x = 66.71 or x = 15.29
The smaller value of x is x₁ = 15.3 ft when the rock emerges out of the hole and reaches ground level.
The larger value of x is x₂ = 66.7 ft when the rock falls to the ground.
The rock lands 66.7 ft horizontally from Landon.
Answer: 66.7 ft
The path of the rock is
y = -0.005x² + 0.4x - 5.1
When x = 0, y = -5.1 ft, which is the location of the hole level.
The rock reaches ground level two times when x = x₁ and when x = x₂.
At ground level, y = 0. Therefore
-0.005x² + 0.4x - 5.1 = 0
Divide through by -0.005.
x² - 82x + 1020 = 0
Solve with the quadratic formula.
x = (82 +/- √(82² - 4*1020))/2
= (82 +/- 51.4198)/2
x = 66.71 or x = 15.29
The smaller value of x is x₁ = 15.3 ft when the rock emerges out of the hole and reaches ground level.
The larger value of x is x₂ = 66.7 ft when the rock falls to the ground.
The rock lands 66.7 ft horizontally from Landon.
Answer: 66.7 ft