Landon is standing in a hole that is 6.4 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 +
0.49x 6.4, 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? Round your answer to the
nearest hundredth. (Hint: When does y = 0)
Clearly,
y = 0.005x + 0.49x 6.4
is not the function involved. Try typing instead of copy/pasting. I'd guess you meant
y = -0.005x^2 + 0.49x - 6.4
The rock hits the ground when the height is zero, right? Heck, they even gave you the hint.
The only trouble I see is that a projectile thrown at an angle θ from height h has height
y = -g/(2 (v cosθ)^2) x^2 + x tanθ + h
Your function would indicate that tanθ = 0.49, so the rock was thrown at an angle of 26°. That's a mighty shallow angle for someone trying to throw a rock from 6 ft under!