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.
(1 point)
Responses

4.30 m
4.30 m

160.56 m
160.56 m

161.12 m
161.12 m

13.94 m

Answer 1

The rocket will land when it reaches a height of 0. To find the horizontal distance at which this occurs, we set y = 0 in the equation:

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

Solving for x using the quadratic formula:

x = (-9.6 ± sqrt(9.6^2 - 4(-0.06)(5.4))) / 2(-0.06)
x = (-9.6 ± sqrt(92.16 + 1.296)) / -0.12
x = (-9.6 ± sqrt(93.456)) / -0.12
x ≈ (-9.6 ± 9.67) / -0.12

Taking the positive value since negative distance doesn't make sense:

x ≈ (0.07) / -0.12
x ≈ -0.5833 m

Rounded to the nearest hundredth, the rocket will land approximately 0.58 m horizontally from its starting point. None of the given responses match this result, so there may have been a mistake in the calculations.