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.

A. 4.30 m

B. 160.56 m

C. 161.12 m

D. 13.94 m

Answer 1

To find out how far horizontally from its starting point the rocket will land, we need to determine when the height y of the rocket is 0 (which means it has landed).

Setting y = 0 in the equation, we get:
0 = -0.06x^2 + 9.6x + 5.4

Solving this quadratic equation, we get:
-0.06x^2 + 9.6x + 5.4 = 0
Dividing by -0.06 to simplify:
x^2 - 160x - 90 = 0

Using the quadratic formula, we get:
x = (160 ± sqrt((160)^2 - 4(1)(-90))) / 2(1)
x = (160 ± sqrt(25600 + 360))/2
x = (160 ± sqrt(25960))/2
x = (160 ± 161.12)/2

Therefore, x = 320.12/2 or -1.12/2

So, x = 160.06 m or -0.56 m

Since the question asks only for the positive distance, the rocket will land approximately 160.56 m horizontally from its starting point, which is answer B.