Question

A firefighter holds a hose 2m off the ground and directs a stream of water toward a burning building. The water leaves the hose at an initial speed of 14m /sec at an angle of 30°. The height of the water can be approximated by h(x)=-0.027x²+0.597x+2, where h(x) is the height of the water in meters at a point x meters horizontally from the firefighter to the building.

Part 1 

a. Determine the horizontal distance from the firefighter at which the maximum height of the water occurs.

The water reaches a maximum height when the horizontal distance from the firefighter to the building is approximately _______ m. Round to 1 decimal place.

part 2 

what is the height of the water?

Answer 1

Where h(x) is the height of water and x is the horizontal distance from firefighter to the building.

We need to find x.

Therefore equate h(x)=0 to obtain the value of x.

$\Rightarrow h(x)=-0.027x^2+0.597x+2=0$

$\Rightarrow -0.027x^2+0.597x+2=0$

$\Rightarrow x=\frac{-0.597\pm \sqrt{(0.597)^2-[4*(-0.027)*(2)]}}{2*(-0.027)}$

$\Rightarrow x=\frac{-0.597\pm \sqrt{0.3564-0.216}}{-0.054}$

$\Rightarrow x=\frac{-0.597\pm \sqrt{0.572}}{-0.054}$

$\Rightarrow x=\frac{-0.597\pm 0.756}{-0.054}$

$\Rightarrow x=\frac{-0.597+ 0.756}{-0.054}\;\;\;\;\;\;,\;\;\;\;x=\frac{-0.597- 0.756}{-0.054}$

$\Rightarrow x=\frac{0.15}{-0.054}\;\;\;\;\;\;,\;\;\;\;x=\frac{-0.353}{-0.054}$

$\Rightarrow x=-2.955\;\;\;\;\;\;,\;\;\;\;x=25.05$

x is in metres and distance cant be negative

$\Rightarrow x=25.05$

Part 2

We know that for a quadratic function , maximum occurs at x = -b / 2a if a < 0

In this case, a = -0.027 and b = 0.556

Since a < 0, therefore maximum height of the water is when

x = -b/2a = - 0.556 / 2(-0.027) = 10.296

Substituting x = 10.296 we get

Maximum height of the water

= 7.862 m