To find the horizontal distance at which the rock will land, we need to determine the value of x when y is equal to 0. In other words, we need to find the x-intercepts of the equation.
Setting y to 0, we have:
0 = -0.005x^2 + 0.48x - 6.7
Now, let's solve this quadratic equation. There are multiple ways to solve it, but let me show you one method using the quadratic formula:
The quadratic formula is given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, the coefficients of the quadratic equation are as follows:
a = -0.005
b = 0.48
c = -6.7
Plugging these values into the quadratic formula, we get:
x = (-0.48 ± √(0.48^2 - 4(-0.005)(-6.7))) / (2(-0.005))
Simplifying further:
x = (-0.48 ± √(0.2304 - 0.134)) / (-0.01)
x = (-0.48 ± √0.0964) / (-0.01)
We have two potential solutions, one with the plus sign (+) and one with the minus sign (-). Let's calculate both of them:
x1 = (-0.48 + √0.0964) / (-0.01) ≈ 17.27563
x2 = (-0.48 - √0.0964) / (-0.01) ≈ 79.72437
Now, considering the context of the problem, we can determine that Landon is standing at the origin (0,0). Therefore, the horizontal distance at which the rock will land from Landon is given by the value of x.
From our calculations, we have two possible distances: approximately 17.28 meters and approximately 79.72 meters. However, since the rock was thrown upwards, it is more reasonable to consider the positive distance, 17.28 meters, as the answer.
Therefore, the rock will land approximately 17.28 meters horizontally from Landon's position.