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 0.06 x squared plus 9.6 x plus 5.4 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.

Question

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 0.06 x squared plus 9.6 x plus 5.4 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.

Answer 1

To find where the rocket lands, we need to find the x-coordinate where y = 0 (the height above the ground is 0).

-0.06x^2 + 9.6x + 5.4 = 0

Now we can solve for x using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -0.06, b = 9.6, and c = 5.4. Plugging in these values:

x = (-9.6 ± √(9.6^2 - 4(-0.06)(5.4))) / 2(-0.06)
x = (-9.6 ± √(92.16 + 1.296)) / -0.12
x = (-9.6 ± √93.456) / -0.12

x ≈ (-9.6 ± 9.65) / -0.12

x ≈ (-19.25) / -0.12 or x ≈ 0.65 / -0.12

x ≈ 160.41 or x ≈ -5.42

Since the rocket cannot land at a negative distance from its starting point, the rocket will land approximately 160.41 meters horizontally from its starting point.