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

To find where the rocket will land, we need to find the x-value when y equals 0 (the rocket has landed on the ground).

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

Solve for x using the quadratic formula:
x = (-9.6 ± √(9.6^2 - 4*(-0.06)*5.4)) / (2*(-0.06))

x = (-9.6 ± √(92.16 + 1.3)) / (-0.12)

x = (-9.6 ± √93.46) / (-0.12)

x ≈ (-9.6 ± 9.667) / (-0.12)

x ≈ (-19.267) / (-0.12) or x ≈ (0.67) / (-0.12)

x ≈ 160.56 m or x ≈ -5.58 m

Since the rocket is landing in the field, the positive value is the distance from its starting point where it will land. Therefore, the rocket will land approximately 160.56 meters horizontally from its starting point.

So, the answer is: 160.56 m