A fire hose ejects a stream of water at an angle of 31.7 ° above the horizontal. The water leaves the nozzle with a speed of 29.2 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?

Vo = 29.2m/s[31.7o]

Xo = 29.2*Cos31.7 = 24.84 m/s
Yo = 29.2*sin31.7 = 15.34 m/s

Y = Yo + g*Tr = o
Tr = -Yo/g = -15.34/-9.8 = 1.57 s. = Rise time.

Dx = Xo*Tr = 24.84m/s * 1.57s = 38.9 m.

To determine how far from the building the fire hose should be located to hit the highest possible fire, we need to find the horizontal distance traveled by the water stream.

The initial speed of the water stream is 29.2 m/s, and it is ejected at an angle of 31.7° above the horizontal. We can split the initial velocity into horizontal and vertical components:

The horizontal component can be found using the equation:
Vx = V * cos(θ)
where V is the initial velocity and θ is the angle.

Vx = 29.2 m/s * cos(31.7°)
Vx = 29.2 m/s * 0.857
Vx ≈ 25.02 m/s

The vertical component can be found using the equation:
Vy = V * sin(θ)
where V is the initial velocity and θ is the angle.

Vy = 29.2 m/s * sin(31.7°)
Vy = 29.2 m/s * 0.528
Vy ≈ 15.39 m/s

The time it takes for the water stream to reach its highest point can be found using the vertical component of the initial velocity and the acceleration due to gravity (g = 9.8 m/s^2).

Using the equation:
t = Vy / g

t = 15.39 m/s / 9.8 m/s^2
t ≈ 1.57 s

The horizontal distance traveled by the water stream can be found using the horizontal component of velocity and the time it takes to reach the highest point:

Distance = Vx * t

Distance = 25.02 m/s * 1.57 s
Distance ≈ 39.28 meters

Therefore, the fire hose should be located approximately 39.28 meters away from the building to hit the highest possible fire.

To determine how far from the building the fire hose should be located to hit the highest possible fire, we need to analyze the motion of the water as a projectile.

Let's break down the information given:
- The water leaves the nozzle with a speed of 29.2 m/s.
- The angle above the horizontal at which the water is ejected is 31.7°.

First, we need to find the time it takes for the water to reach its highest point. At the highest point of the projectile motion, the vertical component of velocity is zero. We can use the following equation to find the time of flight:

v = u + at,

where,
v = final velocity (0 m/s at the highest point),
u = initial velocity (29.2 m/s, the water's speed),
a = acceleration (-9.8 m/s^2, due to gravity),
t = time of flight (unknown).

Rearranging the equation to solve for time (t), we get:

t = (v - u) / a.

Substituting the given values:

t = (0 - 29.2) / -9.8.

Calculating this, we find that t ≈ 2.98 seconds.

Since the water reaches its highest point halfway through its total time of flight, the time it takes to reach its highest point is t/2 ≈ 1.49 seconds.

Next, we can find the horizontal distance traveled by the water during this time. The horizontal component of velocity remains constant throughout the motion. We can use the following equation to calculate the distance:

s = ut,

where,
s = horizontal distance (unknown),
u = horizontal component of velocity (u * cosθ, where θ is the angle of projection),
t = time of flight (1.49 seconds).

Substituting the known values:

s = (29.2 m/s) * cos(31.7°) * (1.49 s).

Calculating this, we find that s ≈ 37.6 meters.

Therefore, the fire hose should be located approximately 37.6 meters away from the building to hit the highest possible fire.