To find the horizontal distance from Landon where the rock will land, we need to determine the value of x when y becomes equal to zero. This is because when the rock lands, its height above the ground will be zero.
Given the equation y = -0.005x^2 + 0.45x - 6.5, we can set y to zero and solve for x:
0 = -0.005x^2 + 0.45x - 6.5
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -0.005, b = 0.45, and c = -6.5. Substituting these values into the formula:
x = (-(0.45) ± √((0.45)^2 - 4(-0.005)(-6.5))) / (2(-0.005))
Simplifying further:
x = (-0.45 ± √(0.2025 - 0.13)) / (-0.01)
x = (-0.45 ± √0.0725) / (-0.01)
x = (-0.45 ± 0.269) / (-0.01)
Now, we have two possible values for x, obtained by both adding and subtracting 0.269:
x1 = (-0.45 + 0.269) / (-0.01)
x1 = -0.181 / (-0.01)
x1 = 18.1
x2 = (-0.45 - 0.269) / (-0.01)
x2 = -0.719 / (-0.01)
x2 = 71.9
Since we're dealing with the horizontal distance, the negative value (x1) doesn't make sense in this context. Therefore, the rock will land approximately 71.9 meters horizontally from Landon.