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 = –0.02x2 + 2.3x + 6, 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?

Answer 1

To find the horizontal distance at which the rocket lands, we need to find the value of x when y is equal to zero.

So, we set the equation equal to zero and solve for x:

0 = -0.02x^2 + 2.3x + 6

We can try factoring, but since the coefficient of x^2 is not 1, factoring might not work easily.

Alternatively, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

In this case, a = -0.02, b = 2.3, and c = 6.

x = (-2.3 ± sqrt(2.3^2 - 4(-0.02)(6))) / (2(-0.02))

Calculating the discriminant:

b^2 - 4ac = 2.3^2 - 4(-0.02)(6) = 5.29 + 0.48 = 5.77

Plugging into the quadratic formula:

x = (-2.3 ± sqrt(5.77)) / (-0.04)

The square root of 5.77 is approximately 2.402.

x = (-2.3 ± 2.402) / (-0.04)

Now, let's calculate both solutions:

x1 = (-2.3 + 2.402) / (-0.04) ≈ 25.5 meters
x2 = (-2.3 - 2.402) / (-0.04) ≈ -12.55 meters

Since distance cannot be negative, the rocket lands approximately 25.5 meters horizontally from its starting point.