A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y= -0.06x^2 + 9.6 x + 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.

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= -0.06x^2 + 9.6 x + 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.

• 4.30 m
• 160.56 m
• 161.12 m
• 13.94 m

Answer 1

With a tricky question like this (that does not factor easily), use the quadratic formula to solve for the zeros (the place the rocket hits the ground)

a = -.06
b = 9.6
c = 5.4
x = (-b + or - the square root of (b^2-4ac)) all divided by (2a)

Answer 2

To find the horizontal distance at which the rocket will land, we need to find the x-coordinate when y is equal to zero.

Given the equation: y = -0.06x^2 + 9.6x + 5.4

Setting y = 0:
0 = -0.06x^2 + 9.6x + 5.4

Now, we can solve this quadratic equation. There are multiple methods to solve it, but one way is to use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = -0.06, b = 9.6, and c = 5.4.

Plugging in the values:
x = (-9.6 ± √(9.6^2 - 4 * -0.06 * 5.4)) / (2 * -0.06)

Simplifying further:
x = (-9.6 ± √(92.16 + 1.296)) / (-0.12)
x = (-9.6 ± √93.456) / -0.12

Taking the positive value:
x = (-9.6 + √93.456) / -0.12
x ≈ -9.6 + 9.67 / -0.12
x ≈ -0.07 / -0.12
x ≈ 0.58

So, the rocket will land approximately 0.58 meters horizontally from its starting point.

Rounded to the nearest hundredth, the answer is 0.58 m.

Answer 3

To find the horizontal distance at which the rocket will land, we need to determine the value of x when the height (y) is equal to zero. This is because when the rocket lands, its height above the ground will be zero.

So we need to solve the equation -0.06x^2 + 9.6x + 5.4 = 0 for x.

To solve this quadratic equation, we can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients in the equation of the curve.

In this case, the equation is -0.06x^2 + 9.6x + 5.4 = 0, so a = -0.06, b = 9.6, and c = 5.4.

Plugging these values into the quadratic formula, we get:
x = (-9.6 ± sqrt(9.6^2 - 4*(-0.06)*5.4)) / (2*(-0.06))

Simplifying further:
x = (-9.6 ± sqrt(92.16 + 1.296)) / (-0.12)
x = (-9.6 ± sqrt(93.456)) / (-0.12)

Calculating the square root: sqrt(93.456) ≈ 9.6724
x = (-9.6 ± 9.6724) / (-0.12)

Now we have two possible values for x:
x1 = (-9.6 + 9.6724) / (-0.12)
x2 = (-9.6 - 9.6724) / (-0.12)

Calculating:
x1 ≈ -0.0724 / (-0.12) ≈ 0.603; round to two decimal places: x1 ≈ 0.60 m
x2 ≈ -19.2724 / (-0.12) ≈ 160.604; round to two decimal places: x2 ≈ 160.60 m

Since we are looking for the positive distance, the rocket will land approximately 160.60 meters horizontally from its starting point. Therefore, the answer is:

• 160.56 m