The answer is "parabola [tex]y =-0.0139(x-60)^2+50[/tex]", 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[tex]y = a(x-h)^2+k[/tex]
- [tex](h,k)=(60, 50)[/tex] the equation will thus be [tex]y=a(x-60)^2+50[/tex]
- Since we assume, the graph, therefore, crosses from point [tex](0,0)[/tex]
[tex]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[/tex]
Therefore, the equation of the given parabola is [tex]y =-0.0139(x-60)^2+50[/tex]
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