To find out where the rocket will land, we need to find the value of x when y is equal to 0 (since the rocket will be at ground level).
Setting y equal to 0 in the equation, we get:
0 = -0.02x^2 + 2.3x + 6
To solve this quadratic equation, we can use 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 quadratic formula:
x = (-2.3 ± sqrt(2.3^2 - 4(-0.02)(6))) / (2 * -0.02)
x = (-2.3 ± sqrt(5.29 + 0.48)) / (-0.04)
x = (-2.3 ± sqrt(5.77)) / (-0.04)
Using a calculator, we can find the square root of 5.77 to be approximately 2.40:
x = (-2.3 ± 2.40) / (-0.04)
Considering both positive and negative solutions, we have:
x = (-2.3 + 2.40) / (-0.04) ≈ 2.50 / (-0.04) = -62.5
x = (-2.3 - 2.40) / (-0.04) ≈ -4.70 / (-0.04) = 117.50
Since distance cannot be negative, we disregard the negative solution.
So, the rocket will land approximately 117.50 meters horizontally from its starting point.
Therefore, the correct answer is option 3) 117.55 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 57.50 m
2 115.00 m
3 117.55 m
4 235.10 m
1 answer