A fireman, 45.1 m away from a burning building, directs a stream jet of water from a ground level fire hose at an angle of 34.1 degrees above the horizontal.

The acceleration of gravity is 9.8 m/s^2.
If the speed of the stream as it leaves the hose is 58.8 m/s, at what height will the stream of water strike the building?

Answer is: 26.3309 m

Not sure how to get the answer; how do I know when the water is headed back down? Is there a way to integrate velocity and find where the slope is equal to zero? If so, is velocity a function of time or distance here? If someone could give me the general equation and show me how to apply it that would be great.

First figure out how long it take the water to reach the wall of the building.

That time T is 45.1 m/(58.8 cos34.1 m/s) T = 0.926269 s

Use the equation for vertical height vs time to determine where it is when it hits the building.

Y = Vo*sin34.1*T - (1/2)gT^2
= 58.8*0.56064*0.926269 s
- (1/2)(9.80)(0.926269)
= 30.535 - 4.543 = 26.0 m

There are too many significant figures in your answer. You did not mention the elevation of the hose where the water left it. That would have to be added to the result.

To solve this problem, we can analyze the vertical and horizontal components of the water stream separately. Let's start by finding the time it takes for the water to reach the building.

We can use the horizontal component of the stream's velocity to determine the time it takes to reach the building. Since there is no horizontal acceleration, the horizontal velocity remains constant. The horizontal velocity can be found using the equation:

vx = v * cos(θ)

where:
vx = horizontal velocity
v = speed of the stream = 58.8 m/s
θ = angle of the stream above the horizontal = 34.1 degrees

Now, let's calculate the horizontal velocity:

vx = 58.8 * cos(34.1) = 48.9685 m/s

Next, we can find the time it takes for the water to reach the building by dividing the horizontal distance by the horizontal velocity:

t = distance / vx

distance = 45.1 m
vx = 48.9685 m/s

t = 45.1 / 48.9685 = 0.9204 s

Now that we know it takes approximately 0.9204 seconds for the water to reach the building horizontally, we can find the height at which it strikes the building.

The vertical motion of the water can be described by the equation:

y = yo + voy * t - (1/2) * g * t^2

where:
y = final height (what we want to find)
yo = initial vertical position (ground level)
voy = initial vertical velocity (vertical component of the stream's velocity)
g = acceleration due to gravity = 9.8 m/s^2
t = time = 0.9204 s

At the highest point of the water stream, the vertical velocity would be momentarily zero. So, we need to find the vertical velocity (voy) at time t = 0.9204 seconds.

The vertical velocity can be found using the equation:

voy = v * sin(θ)

where:
voy = vertical component of the stream's velocity
v = speed of the stream = 58.8 m/s
θ = angle of the stream above the horizontal = 34.1 degrees

Now, let's calculate the vertical velocity:

voy = 58.8 * sin(34.1) = 32.9141 m/s

Substituting the values into the initial equation for vertical motion, we can solve for the final height (y):

y = 0 + 32.9141 * 0.9204 - (1/2) * 9.8 * (0.9204)^2
y ≈ 26.3309 m

Therefore, the stream of water will strike the building at a height of approximately 26.3309 meters.