Landon is standing in a hole that is 5.1ft deep. He threw 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

anonymous · Answer

Since you're looking for the horizontal position of the rock when the rock is on the ground, you're looking for a solution where y=0 (since y is the vertical position of the rock). So we set y=0 and solve the equation, which gives us two results. One is about 15.3, the other about 66.7. Intuitively, we know that the rock will be at y=0 once on its way up, and then a second time on its way back down, when it lands. Since we're looking for its horizontal distance when it lands, we're looking for the larger solution, which is roughly 66.7 feet.