To find where the rocket will land, we need to look for the x-coordinate when y = 0 (since the rocket will be on the ground).
0 = -0.04x^2 + 8.3x + 4.3
Multiplying by 100 to make the numbers easier to work with:
0 = -4x^2 + 830x + 430
Now we need to solve this quadratic equation. We can use the quadratic formula:
x = (-b ± √(b^2-4ac)) / 2a
Plugging in a = -4, b = 830, and c = 430:
x = (-830 ± √(830^2 - 4*(-4)*430)) / (2*(-4))
x = (-830 ± √(688900 - 6880)) / -8
x = (-830 ± √682020) / -8
x = (-830 ± 825.31) / -8
Taking the positive root (x must be positive since it's measuring distance):
x = (-830 + 825.31) / -8
x = -4.69 / -8
x = 0.58625
Converting back to meters, we get 0.59 m. Therefore, the rocket will land approximately 0.59 m horizontally from its starting point.
So the closest answer choice is:
0.52 m
A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.04x2+ 8.3x + 4.3, 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?
(1 point)
Responses
208.02 m
208.02 m
416.03 m
416.03 m
0.52 m
0.52 m
208.19 m
208.19 m
1 answer