A fireman d = 56.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of θi = 24.0° above the horizontal as shown in the figure. If the speed of the stream as it leaves the hose is vi = 40.0 m/s, at what height will the stream of water strike the building?
I will be happy to critique your work.
From the x direction:
t = 56/40cos24
Use this t in the y equation:
y = 40sin24 t - .5*9.8*t^2
To find the height at which the stream of water will strike the building, we need to break down the problem into horizontal and vertical components.
First, let's calculate the time it takes for the water stream to reach the building. We can use the horizontal component of the velocity, vix, and the distance, d, to find the time, t.
Horizontal component of velocity, vix = vi * cos(θi)
vix = 40.0 m/s * cos(24.0°)
Next, calculate the time, t:
t = d / vix
t = 56.0 m / (40.0 m/s * cos(24.0°))
Now, let's calculate the vertical distance the water stream will travel in time t. We can use the vertical component of the velocity, viy, and the time, t, to find the vertical distance, h.
Vertical component of velocity, viy = vi * sin(θi)
viy = 40.0 m/s * sin(24.0°)
Now, calculate the vertical distance, h:
h = viy * t
h = (40.0 m/s * sin(24.0°)) * (56.0 m / (40.0 m/s * cos(24.0°)))
After calculating the above expression, we will get the vertical distance, which is the height at which the stream of water will strike the building.