11. 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)
57.50 m
115.00 m
117.55 m
235.10 m
To find where the rocket lands, we need to find the x-intercept of the equation (where y equals 0).
0 = -0.02x^2 + 2.3x + 6
We can solve for x using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = -0.02, b = 2.3, and c = 6. Plugging these values into the formula, we get:
x = (-2.3 ± sqrt(2.3^2 - 4(-0.02)(6))) / 2(-0.02)
Simplifying:
x = (-2.3 ± sqrt(54.49)) / (-0.04)
x = (-2.3 ± 7.38) / (-0.04)
x = 57.5 or x = 117.55
The rocket will land either 57.5 meters or 117.55 meters horizontally from its starting point.
Answer: C) 117.55 m
that makes so much sense
I'm glad I could help clarify the solution! Do you have any other questions or is there anything else I can assist you with?
To find the horizontal distance from the starting point where the rocket will land, we need to find the x-value when y = 0.
The equation is y = -0.02x^2 + 2.3x + 6.
Setting y to 0, we get:
0 = -0.02x^2 + 2.3x + 6
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -0.02, b = 2.3, c = 6.
Plugging in these values, we get:
x = (-2.3 ± √(2.3^2 - 4(-0.02)(6))) / (2(-0.02))
Simplifying further:
x = (-2.3 ± √(5.29 + 0.48)) / -0.04
x = (-2.3 ± √5.77) / -0.04
Using a calculator, we find that the approximate values are:
x = (-2.3 + √5.77) / -0.04 ≈ 57.50
x = (-2.3 - √5.77) / -0.04 ≈ 235.10
Since the rocket is launched horizontally, we ignore the negative value. Therefore, the rocket will land approximately 57.50 meters horizontally from its starting point.
Therefore, the correct answer is: 57.50 m.