Question

A fireboat in the harbor is helping put out a fire in a warehouse on the pier. The
distance from the barrel (end) of the water cannon to the roof of the warehouse is
120 feet, and the water shoot up 50 feet above the barrel of the water cannon.
Write a function that models the path of the water.

Answer 1

The answer is "parabola $y =-0.0139(x-60)^2+50$", and the further calculation can be defined as follows:

• Let's all describe the circumstance in an XY plane and then let the origin (0,0) indicate the parabola's initial point.
• The last point now, so because the diameter of the parabola is 120 feet (120, 0).
• It is stated that, as a result of the water shots at 50 feet just above the cannon barrel, the vertex of such a parabola shall be, as shown in the picture below, (60, 50) as a parabola is a symmetric figure.
• Now, we know (h,k) that the general parabolic equation$y = a(x-h)^2+k$
• $(h,k)=(60, 50)$ the equation will thus be $y=a(x-60)^2+50$
• Since we assume, the graph, therefore, crosses from point $(0,0)$

          $y =a(x-60)^2+50\\\\0=a(0-60)^2 + 50\\\\0 = 3600a + 50 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(subtract 50 from both sides)}\\\\0 - 50 = 3600a + 50 - 50\\\\-50=3600a \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(divide both sides by 3600)}\\\\\frac{-50}{3600}=\frac{3600 a}{3600} \\\\-0.0139 \approx a \\\\a \approx -0.0139$

Therefore, the equation of the given parabola is $y =-0.0139(x-60)^2+50$

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