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Exercise 12.3 · Q5

Q.A water jet from a fountain reaches its maximum height of 4 metres at a distance of 0.5 metres from the vertical passing through the point O of the water outlet. Find the height of the jet above the horizontal OX at a distance 0.75 metre from the point O.

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Model the water jet as a downward-opening parabola with vertex at the point of maximum height; use the fact that it passes through the origin O to fix the parabola, then evaluate at x=0.75x=0.75 m.

Parabola with vertex (h,k)(h,k), axis vertical, opening downward: (x−h)2=−4a(y−k)(x-h)^2=-4a(y-k).

  1. The jet reaches maximum height 4 m at horizontal distance 0.5 m from O, so the vertex is at (h,k)=(0.5,4)(h,k)=(0.5,4).
  2. Parabola: (x−0.5)2=−4a(y−4)(x-0.5)^2=-4a(y-4).
  3. The jet starts at the outlet O =(0,0)=(0,0), so this point lies on the parabola. Substitute (0,0)(0,0):

(0−0.5)2=−4a(0−4)⇒0.25=16a⇒a=0.2516=164(0-0.5)^2=-4a(0-4)\Rightarrow0.25=16a\Rightarrow a=\frac{0.25}{16}=\frac{1}{64}

  1. Equation: (x−0.5)2=−4(164)(y−4)=−116(y−4)(x-0.5)^2=-4\left(\dfrac{1}{64}\right)(y-4)=-\dfrac{1}{16}(y-4).
  2. Substitute x=0.75x=0.75: (0.75−0.5)2=(0.25)2=0.0625(0.75-0.5)^2=(0.25)^2=0.0625. 0.0625=−116(y−4)⇒y−4=−0.0625×16=−1⇒y=4−1=30.0625=-\frac{1}{16}(y-4)\Rightarrow y-4=-0.0625\times16=-1\Rightarrow y=4-1=3 …

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