A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation-
y= -0.04x^2+8.3x,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 ground. How far horizontally from its starting point will the rocket land?
a- 208.02m
b- 416.03m
c- 0.52m
d- 208.19m
This 68 year old grandfather is lost on this one.
y = x (-.04 x + 8.3)
when is y = 0
at the start of course but also when
.04 x = 8.3
or x = 207.5 meters
I think you left a significant figure out :)
your right sorry
y=-0.04x^2+8.3x+4.3
.04 x^2 - 8.3 x - 4.3 = 0
http://www.webgraphing.com/quadraticequation_quadraticformula.jsp#
To find out how far horizontally from its starting point the rocket will land, we need to determine the x-coordinate when the height of the rocket, represented by y, is equal to zero. This is because y represents the height above ground, and when the rocket lands, its height will be zero.
To find the x-coordinate when y = 0, we set the equation:
-0.04x^2 + 8.3x = 0
Next, we can factor out an x:
x(-0.04x + 8.3) = 0
Now we have two possible solutions:
1) x = 0 (represents starting point on the roof)
2) -0.04x + 8.3 = 0
Solving the second equation:
-0.04x + 8.3 = 0
-0.04x = -8.3
x = -8.3 / -0.04
x ≈ 207.5
Since the rocket is launched horizontally, it cannot land behind its starting point, so we disregard the x = 0 solution. Therefore, the rocket will land approximately 207.5 meters horizontally from its starting point.
Among the answer choices provided, the closest option to 207.5 meters is d) 208.19m.