A physics student stands at the top of a hill that has an elevation of 37 meters. He throws a rock and it goes up into the air and then falls back past him and lands on the ground below. the path of the rock can be modeled by the equation y = -0.2x^2 + 0.8x +37 where x is the horizontal distance, in meters, from the starting point on the top of the hill and y is the height, in meters, of the rock above the ground. How far horizontally from its starting point will the rock land?

Question

A physics student stands at the top of a hill that has an elevation of 37 meters. He throws a rock and it goes up into the air and then falls back past him and lands on the ground below. the path of the rock can be modeled by the equation y = -0.2x^2 + 0.8x +37 where x is the horizontal distance, in meters, from the starting point on the top of the hill and y is the height, in meters, of the rock above the ground. How far horizontally from its starting point will the rock land?
a. 37.00
b. 67.43***
c. 27.43
d. 37.78

Anonymous · Answer

X = (-0.8 +- sqrt(0.64+2.96))/-0.04 = -27.43, and 67.43 X = 67.43

Henry · Answer

-0.02x^2 Not -0.2x^2. Your answer is correct.