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?

Vi=29.2 m/s

θ=31.7°
Vxi=Vi cos(θ) = 24.844 m/s
Vyi=Vi sin(θ) = 15.344 m/s
since
Vy(t)=Vyi-gt
so at the maximum,
Vy(t)=0, or Vyi-gt=0
=>
t=Vyi/g
The optimal (best) horizontal distance would therefore be the horizontal distance travelled during this time interval, or
X = (Vyi/g)*Vxi
(for checking: 35<X<40 m)

To solve this problem, we need to analyze the motion of the water as a projectile and find the horizontal distance (range) it will travel before hitting the ground. In this case, the highest possible fire is reached when the highest point of the water's trajectory intersects with the height of the fire.

Let's break down the problem into components:

1. Resolve the initial velocity into horizontal and vertical components:
The vertical component is given by: V_y = V * sinθ
The horizontal component is given by: V_x = V * cosθ, where V is the initial speed (29.2 m/s) and θ is the launch angle (31.7°).

2. Find the time it takes for the water to reach its highest point:
The time taken to reach the highest point can be found using the equation: t = V_y / g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).

3. Find the maximum height reached by the water:
The maximum height can be calculated using the equation: H = (V_y^2) / (2g).

4. Determine the range of the water:
The total time of flight can be calculated using the equation: T = 2t.
The range (R) can be found using the equation: R = V_x * T.

By using these steps, we can find the distance from the building where the fire hose should be located to hit the highest possible fire.