Question
The underside of a concrete underpass forms a parabolic arch. The arch is 40 m wide at the base and 11.2 m high in the centre. The upper surface of the underpass is 60 m wide and the concrete is 2.5 m thick at the centre. What would be the minimum headroom on a sidewalk that is built 1.8 m from the base of the underpass?
Answers
Don't you want the maximum headroom?
for the underside, vertex is (0,11.2)
equation:
y = ax^2 + 11.2
but (20,0) lies on it
0 = a(400) + 11.2
a = -11.2/400 = -112/4000 = -7/250
equation: y = -7/250 x^2 + 11.2
so 1.8 m from base ---> x = 18.2
y = -7/250(18.2^2) + 11.2 = 1.93 m
What does the upper surface have to do with this?
for the underside, vertex is (0,11.2)
equation:
y = ax^2 + 11.2
but (20,0) lies on it
0 = a(400) + 11.2
a = -11.2/400 = -112/4000 = -7/250
equation: y = -7/250 x^2 + 11.2
so 1.8 m from base ---> x = 18.2
y = -7/250(18.2^2) + 11.2 = 1.93 m
What does the upper surface have to do with this?
I think they mean minimum required headroom
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