SOLUTION: a parkway 80 ft. wide is spanned by a parabolic arc 100 ft. long along the horizontal, if the parkway is in the center, how high must the vertex of the arch be in order to give a m

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: a parkway 80 ft. wide is spanned by a parabolic arc 100 ft. long along the horizontal, if the parkway is in the center, how high must the vertex of the arch be in order to give a m      Log On


   

Question 1005396: a parkway 80 ft. wide is spanned by a parabolic arc 100 ft. long along the horizontal, if the parkway is in the center, how high must the vertex of the arch be in order to give a minimum clearance of 20 over the parkway?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The parabola needed, drawn over a conveniently placed set of axes, with all measures in feet, looks like this:
The equation will be y=f%28x%29=ax%5E2%2Bk such that system%28f%2840%29=20%2Cf%2850%29=0%29 .
system%28y=f%28x%29=ax%5E2%2Bk%2Cf%2840%29=20%2Cf%2850%29=0%29--->system%28y=f%28x%29=ax%5E2%2Bk%2Ca40%5E2%2Bk=20%2Ca50%5E2%2Bk=20%29--->system%28y=f%28x%29=ax%5E2%2Bk%2C1600a%2Bk=20%2C2500a%2Bk=0%29--->system%28y=f%28x%29=-x%5E2%2F45%2B500%2F9%2Ca=-1%2F45%2Ck=500%2F9%29
The height at the vertex (in feet) is f%280%29=highlight%28500%2F9=%2255.55+%28+rounded+%29%22%29 .