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Exercise 5.5 · Q1

Q.A bridge has a parabolic arch that is 10 m10\,\text m high in the centre and 30 m30\,\text m wide at the bottom. Find the height of the arch 6 m6\,\text m from the centre, on either sides.

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✓ Free question

Place the vertex (the arch's highest point) at the origin of a local coordinate system, use the known base width to find aa, then substitute x=6x=6.

Step 1. Set up coordinates. Put the vertex at (0,10)(0,10) (height 10 m10\,\text m at the centre), parabola opening downward: (x−0)2=−4a(y−10)(x-0)^2=-4a(y-10).

Step 2. Use the base data to find aa. Width 30 m30\,\text m at the bottom (ground level, y=0y=0) means the base corners are (±15,0)(\pm15,0). Substituting (15,0)(15,0):

152=−4a(0−10)⇒225=40a⇒a=22540=45815^2=-4a(0-10) \Rightarrow 225=40a \Rightarrow a=\dfrac{225}{40}=\dfrac{45}8.

Step 3. Find the height at x=6x=6.

62=−4a(y−10)⇒36=−4(458)(y−10)⇒36=−452(y−10)⇒y−10=−7245=−1.6⇒y=8.46^2=-4a(y-10) \Rightarrow 36=-4\left(\dfrac{45}8\right)(y-10) \Rightarrow 36=-\dfrac{45}2(y-10) \Rightarrow y-10=-\dfrac{72}{45}=-1.6 \Rightarrow y=8.4.

Step 4. Interpret. By symmetry, the height is the same, 8.4 m8.4\,\text m, on both sides of the centre.

✓Final answer

8.4 m8.4\,\text m on either side.

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