A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y equals negative 0.8 x squared plus 12 x plus 25.8 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? Round your answer to the nearest hundredth.
A. 25.80 m
B. 37.00 m
C. 17.24 m
D. 16.91 m
To find where the rocket lands, we need to find the value of x when y = 0 (since at the point where the rocket lands, its height above the ground will be 0). So we set the equation equal to 0 and solve for x:
0 = -0.8x^2 + 12x + 25.8
We can simplify this by dividing everything by -0.8:
0 = x^2 - 15x - 32.25
Now we can use the quadratic formula to solve for x:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 1, b = -15, and c = -32.25. Plugging these values in, we get:
x = (-(-15) ± sqrt((-15)^2 - 4(1)(-32.25))) / 2(1)
x = (15 ± sqrt(693)) / 2
We don't need the negative root since that would give us a distance to the left of the starting point, which doesn't make sense. So we take the positive root and round to the nearest hundredth:
x ≈ 16.91
Therefore, the rocket will land about 16.91 meters horizontally from its starting point, which is answer choice D.