The path traveled by a remote controlled model airplane is shaped like a parabola. It took off from the ground and landed on the ground 160 ft. away from where it took off. If it reached a maximum height of 40 ft. write an equation for the parabola that models the path of the plane. Assume the point of take-off is the origin on a coordinate plan.
Thank you.
I appreciate it
Thank you.
I appreciate it
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The equation for a parabola looks like y = a(x - h)^2 + k, where (h,k) is the vertex, and a squashes things to fit.
We know that in this case, a is going to be negative, because the parabola is upside-down.
If the path is parabolic, the vertex is halfway in betweenm, so at 160 ft / 2 = 80 ft.
The vertex is at (80,40), so....
y = a(x - 40)^2 + 80
Now, we have to use a to squish the equation so that at x = 0 or x = 160, y = 0.
0 = a * (0 - 40)^2 + 80
-80 = a * (-40)^2
-80 = a * 1600
-80/1600 = a
a = -1/20
y = (-1/20) * (x - 40)^2 + 80
We know that in this case, a is going to be negative, because the parabola is upside-down.
If the path is parabolic, the vertex is halfway in betweenm, so at 160 ft / 2 = 80 ft.
The vertex is at (80,40), so....
y = a(x - 40)^2 + 80
Now, we have to use a to squish the equation so that at x = 0 or x = 160, y = 0.
0 = a * (0 - 40)^2 + 80
-80 = a * (-40)^2
-80 = a * 1600
-80/1600 = a
a = -1/20
y = (-1/20) * (x - 40)^2 + 80
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h(x) = ax^2 + bx + c
h(0) = 0 so c = 0
h(160) = 0 gives 0 = a(160)^2 + b(160)
h(80) = 40 gives 40 = a(80^2) + b(80)
solve these two equations in two unknowns giving
so b = 1 and a = -1/160
h(x) = -x^2 / 160 + x
checking
h(0) = 0
h(80) = -6400/160 + 80 = 40
h(160) = -25600/160 + 160 = 0
h(0) = 0 so c = 0
h(160) = 0 gives 0 = a(160)^2 + b(160)
h(80) = 40 gives 40 = a(80^2) + b(80)
solve these two equations in two unknowns giving
so b = 1 and a = -1/160
h(x) = -x^2 / 160 + x
checking
h(0) = 0
h(80) = -6400/160 + 80 = 40
h(160) = -25600/160 + 160 = 0