A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.06x2 + 10.1x + 5, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?
To find where the rocket will land horizontally, we need to find the x-coordinate when the rocket hits the ground, which corresponds to the y-coordinate being 0.
So, we can set y = 0 and solve for x:
0 = -0.06x^2 + 10.1x + 5
Multiplying by -1 to make the equation easier to solve gives:
0.06x^2 - 10.1x - 5 = 0
Now we can use the quadratic formula to solve for x:
x = (-(-10.1) ± sqrt((-10.1)^2 - 4(0.06)(-5))) / 2(0.06)
x = (10.1 ± sqrt(102.01 + 1.2)) / 0.12
x = (10.1 ± sqrt(103.21)) / 0.12
x = (10.1 ± 10.16) / 0.12
x = (20.26) / 0.12 or x = (-0.06) / 0.12
x ≈ 168.83 or x ≈ -0.5
Since the rocket cannot land at a negative distance, the rocket will land approximately 168.83 meters horizontally from its starting point.