Question

A tunnel is in the shape of a parabola. The maximum height is 16 m and it is 16 m wide at the base, as shown below. A parabola opening down with vertex at the origin is graphed on the coordinate plane. The height of the parabola from top to bottom is 16 meters and its width from left to right is 16 meters. What is the vertical clearance 7 m from the edge of the tunnel?

Answer 1

The rest of the question are the attached figure and the options to find the answer.

The options are:   A.) 15.4m    B.) 0.3m    C.) 15.7m     D.)0.6m
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Solution:
As shown in the figure, the vertex of the parabola is at the point (0,0)
So, the general equation of that parabola will be :
y = a x²
From the figure we can know that ⇒ at x = 8 → y = -16
∴ -16 = a * 8²
∴ -16 = 64 a
∴ a = -16/64 = -0.25

So, the equation of the figure will be ⇒⇒⇒ y = -0.25 x²

To find the vertical clearance at 7 m from the edge of the tunnel

∴ x = 8 - 7 = 1
By substitute with x=1 at the equation of the figure
∴ y = -0.25 * 1² = -0.25

So, the vertical clearance = 16 - 0.25 = 15.75 m

So, the correct answer is option C.) 15.7m

Answer 2

This a vertical parabola but upside down so x term is squared.

The general equation of a parabola is y=a$(x-h)^{2}$+k

Where, (h,k)= vertex of the parabola

The parabola has vertex at origin, so (h,k) =(0,0)

So, the equation becomes, y=a$x^{2}$

Now, the width is 16 meters, So the x-intercepts must be x=-8 and x=+8

So, the points are (-8,-16) and (+8,-16)

To find the value of 'a' Let us plug x=-8, y=-16 in y=a$x^{2}$

So, We get, -16=a$(-8)^{2}$

-16=a*64

Dividing by 64 on both sides

a=$\frac{-16}{64}$

a=-0.25

So, the equation of parabola becomes

y=-0.25$x^{2}$

We need to find vertical clearance 7m from the edge of the tunnel.

As, edge is at x=8, 7m from the edge is x=8-7=1

So, we need to find vertical clearance, or y, 7m , from the edge of the tunnel, or x=1

Now plugging x=1 in y=-0.25$x^{2}$

y=-0.25$1^{2}$

y=-0.25*1

y=-0.25

Vertical clearance= height-y value=16-0.25=15.75

So, Answer=15.75meters