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?

Question

(1 point)
Responses solve it and tell me why

57.50 m
57.50 m

115.00 m
115.00 m

117.55 m
117.55 m

235.10 m

Answer 1

To find the distance horizontally from its starting point that the rocket will land, we need to find the x-coordinate when y = 0 (height of the rocket above the ground).

So, we substitute y = 0 into the equation:
0 = -0.02x^2 + 2.3x + 6

Now, we need to solve this quadratic equation:
-0.02x^2 + 2.3x + 6 = 0

Multiplying the entire equation by -50 to get rid of the decimal:
x^2 - 115x - 300 = 0

Now we can factor this quadratic equation by looking for two numbers that multiply to -300 and add up to -115:
(x - 120)(x + 5) = 0

This gives us x = 120 or x = -5.

Since the rocket was launched from the roof, we ignore the negative value. Therefore, the rocket will land 120 meters horizontally from its starting point.

Correct answer: 115.00 m