Landon is standing in a hole that is 5.1 ft deep. He throws a rock, and it goes up into the air, out of the hole, and then lands on the ground above. The path of the rock can be modeled by the equation y = - 0.005x 2 + 0.41x - 5.1, where x is the horizontal distance of the rock, in feet, from Landon and y is the height, in feet, of the rock above the ground. How far horizontally from Landon will the rock land?

Question

Casey Barnes

Mathematics

11 January, 17:18

Landon is standing in a hole that is 5.1 ft deep. He throws a rock, and it goes up into the air, out of the hole, and then lands on the ground above. The path of the rock can be modeled by the equation y = - 0.005x 2 + 0.41x - 5.1, where x is the horizontal distance of the rock, in feet, from Landon and y is the height, in feet, of the rock above the ground. How far horizontally from Landon will the rock land?

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Answers (1)

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Shepard · Answer 1

You are given the equation y = - 0.005x^2 + 0.41x - 5.1

with x = the horizontal distance (left/right)

and y = the vertical distance (up/down)

you are trying to find the value of x.

y = 5.1ft. The rock lands on the ground above after being thrown by Landon who is 5.1ft below ground in the hole. So the rock has to be thrown at least 5.1ft to get to the ground above.

Setting y = 5.1, you get

-0.005x^2 + 0.41x - 5.1 = 5.1

move the 5.1 from the right to the left, so you have the equation = 0 and you can solve it like any other quadratic equation.

-0.005x^2 + 0.41x - 10.2 = 0

this requires you to use the quadratic equation but after doing so you find that the values for x are:

and you must take the positive value because distance cannot be negative