To find out how far horizontally from its starting point the rocket will land, we need to find the x-coordinate when the rocket hits the ground, which corresponds to y=0.
Thus, we set y = 0 in the equation y = 0.04x^2 + 8.3x + 4.3:
0 = 0.04x^2 + 8.3x + 4.3
We can solve this equation by either factoring or using the quadratic formula. Let's use the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
In this case:
a = 0.04, b = 8.3, c = 4.3
Plugging these values into the formula:
x = [-8.3 ± sqrt((8.3)^2 - 4(0.04)(4.3))] / (2 * 0.04)
x = [-8.3 ± sqrt(68.89 - 0.688)] / 0.08
x = [-8.3 ± sqrt(68.202)] / 0.08
Now calculate:
x = (-8.3 + sqrt(68.202)) / 0.08 ≈ -0.50 meters
x = (-8.3 - sqrt(68.202)) / 0.08 ≈ -171.25 meters
Since we're only interested in the positive horizontal distance, we discard the negative value. Therefore, the rocket will land approximately 0.50 meters horizontally from its starting point.
A model rocket is launched from a roof into a large field. The path of the rocket can be
modeled by the equation y = 0.04x2+ 8.3x + 4.3, 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?
1 answer
1 answer