To find out how far horizontally from its starting point the rocket will land, we need to find where the height y of the rocket above the ground equals 0. This indicates the rocket has landed.
Setting y = 0 in the equation:
0 = -0.02x^2 + 2.3x + 6
Now we need to solve for x. This is a quadratic equation so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
a = -0.02, b = 2.3, c = 6
x = (-2.3 ± √(2.3^2 - 4*(-0.02)*6)) / 2*(-0.02)
x = (-2.3 ± √(5.29 + 0.48)) / -0.04
x = (-2.3 ± √5.77) / -0.04
x = (-2.3 + √5.77) / -0.04 = -117.55 m
x = (-2.3 - √5.77) / -0.04 = 57.50 m
Since we are looking for a positive distance, the rocket will land approximately 57.50 meters away from its starting point.
Therefore, the correct answer is:
57.50 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 = –0.02x2 + 2.3x + 6, 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
57.50 m
57.50 m
115.00 m
115.00 m
117.55 m
117.55 m
235.10 m
1 answer