A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation 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.

Question

A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation 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.

y=-0.06x^(2)+9.6x+5.4

Answer 1

To find where the rocket will land horizontally, we need to find the x-coordinate where the rocket's height is 0. In other words, we need to find the x-value when y=0.

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

We can solve this equation by factoring, completing the square, or using the quadratic formula. The quadratic formula is probably the easiest method in this case.

x = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = -0.06, b = 9.6, and c = 5.4. Plugging these values into the quadratic formula gives:

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

Now, we calculate the two possible x-values:

x ≈ (9.6 + √93.456) / -0.12
x ≈ (9.6 + 9.67) / -0.12
x ≈ 19.27 / -0.12
x ≈ -160.56

and

x ≈ (9.6 - √93.456) / -0.12
x ≈ (9.6 - 9.67) / -0.12
x ≈ -0.07 / -0.12
x ≈ 0.58

Since the negative value is not relevant in this context, we take the positive value of x = 0.58 meters. The rocket will land approximately 0.58 meters horizontally from its starting point.