Question

Check my answer please
A physics student stands at the top of 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 = –0.02x² + 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? Round your answer to the nearest hundredth
A. 37.00m
B. 67.43m <---
C. 27.43m
D. 37.78m

Answer 1

So y = 0 is at ground level or on top of the hill?

if y=0 at the top of the hill find x when y=-37 m
if y=0 at the bottom of the hill find the second point where y=0

Answer 2

So I got it right correct?

Answer 3

idk i havent calculated anything, which interpretation did you use?

y=0 is where?

Answer 4

I think you're right, i tried using y=0 at the top of the hill and the answer wasn't listed, so you're right.

Answer 5

Rock is on ground when height y = 0, so we need to solve
y = -0.02x^2 + 0.8x + 37 = 0, to find options for horizontal distance x.
This equation may seem more normal if we multiply throuhh by -50.
x^2 - 40x - 1850 = 0 with positive solution selected as
x = 20 + 15√(10) ~ 67.43 meters

Answer 6

My work^!

Answer 7

you're right, good work \(\checkmark\)

Answer 8

Thank you