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

The horizontal distance from the firefighter at which the maximum height of the water occurs is approximately 11.0 meters. The maximum height of the water is approximately 6.31 meters. The horizontal distance between the firefighter and the house when the water hits the ground is approximately 11.0 meters.

Step-by-step explanation:

To find the horizontal distance from the firefighter at which the maximum height of the water occurs, we need to find the x-coordinate of the vertex of the parabolic equation hx) = 0.02612x^2 + 0.577x + 3. The x-coordinate of the vertex can be found using the formula x = -b/(2a), where a, b, and c are the coefficients of the quadratic equation.

Let's substitute the coefficients into the formula:

x = -(0.577)/(2(0.02612))

Simplifying the equation, we get:

x = -0.577/(0.05224)

x ≈ -11.039

Since the distance cannot be negative, we take the absolute value of x, which gives us a horizontal distance of approximately 11.0 meters.

b. The maximum height of the water can be found by substituting the x-coordinate of the vertex into the parabolic equation hx):

hmax) = 0.02612(11.0)^2 + 0.577(11.0) + 3

Simplifying the equation, we get:

hmax) ≈ 6.31 meters.

c. To find the horizontal distance between the firefighter and the house when the water hits the ground, we need to set hx) equal to 0 and solve for x:

0 = 0.02612x^2 + 0.577x + 3

This is a quadratic equation, which can be solved using the quadratic formula. However, since we already know the x-coordinate of the vertex (-11.0), we can assume that the water hits the ground at the same horizontal distance from the firefighter. Therefore, the horizontal distance between the firefighter and the house when the water hits the ground is approximately 11.0 meters.

Answer 2

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 = -
`(b)/(2a)`

From the quadratic function,

h(x) = -0.026x²+ 0.577x + 3

a = -0.026

b = 0.577

Therefore, the horizontal distance =
`-(0.577)/(2* (-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 =
`(-b\pm√(b^2-4ac))/(2a)`

x =
`(0.577\pm √((-0.577)^2-4(0.02612)(3))))/(2(0.02612))`

x =
`(0.577\pm√(0.019489))/(0.05224)`

x =
`(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