Question

A firefighter holds a hose 3 m off the ground and directs a stream of water toward a burning building. The water leaves the hose at an initial speed of 16 m/sec at an angle of 300, The height of the water can be approximated by hx)0.02612 + 0.577x+ 3, where hcx) is the height of the water in meters at a point x meters horizontally from the firefighter to the building.
a. Determine the horizontal distance from the firefighter at which the maximum height of the water occurs Round to 1 decimal place. I decimal place branch of the parabola at a height of 6 m. How far is the

b. What is the maximum height of the water? Round to

c. The flow of water hits the house on the downward firefighter from the house? Round to the nearest meter

Answer 1

Answer:

a). Horizontal distance = 11.1 m

b). Maximum height = 6.2 m

c). Firefighter is 13.7 m from the house

Step-by-step explanation:

Given question is incomplete; find the complete question in the attachment.

Height of the water can be determined by the expression,

h(x) = -0.026x²+ 0.577x + 3

Here x = Horizontal distance of the from the firefighter

a). Since the stream of the water will follow a parabolic path, maximum point of the parabola will be = Vertex of the parabolic path

Horizontal distance from the firefighter at which the water achieves the maximum height = -$\frac{b}{2a}$

From the quadratic function,

h(x) = -0.026x²+ 0.577x + 3

a = -0.026

b = 0.577

Therefore, the horizontal distance = $-\frac{0.577}{2\times (-0.02612)}$ = 11.05 m

                                                         ≈ 11.1 meters

b). By putting x = 11.1 in the quadratic equation,

    h(x) = -0.02612(11.1)²+ 0.577(11.1) + 3

           = -3.2182 + 6.4047 + 3

           = 6.18 m

           ≈ 6.2 m

c). For h(x) = 6 m

    6 = -0.02612x² + 0.577(x) + 3

    0.02612x² - 0.577x + 3 = 0

    From quadratic formula,

    x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

    x = $\frac{0.577\pm \sqrt{(-0.577)^2-4(0.02612)(3))}}{2(0.02612)}$

    x = $\frac{0.577\pm\sqrt{0.019489}}{0.05224}$

    x = $\frac{0.577\pm0.1396}{0.05224}$

    x = 13.7 m, 8.37 m

Therefore, the farthest distance of the firefighter from the house will be 13.7 m