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ffj
Jan 18, 2007, 05:30 PM
If you have the correct answer to this question, please show me how you got it, because I do not get it, so help me out :confused:



A firefighter 50.0m away from a burning building directs a stream of water from a fire hose at an angle 30.0 degrees above the horizontal. If the velocity of the stream is 40.0 m/s, at what height will the stream of water strike the building?

I need help quickly, thank you!

Nosnosna
Jan 18, 2007, 05:59 PM
The first thing you need to do is to decompose the velocity into its vertical and horizontal components. The two components are found using simple trig: a 30-60-90 triangle representing the components. In this case, the velocity is the hypotenuse, the vertical component is the shorter leg (the leg opposite the 30 degree angle), and the horizontal component is the longer leg (opposite the 60 degree angle).

A 30-60-90 triangle always has sides in the following proportions: 1:sqrt(3):2, given in the same order as the angles listed and corresponding to the side opposite the angle. We know that the hypotenuse is 40, so the vertical component is going to be 20 and the horizontal component will be 20*sqrt(3).

The horizontal component is constant:
Vh(t) = 20*sqrt(3)

The vertical component will vary with time, due to gravity. Since gravity is going to have an effect of -9.8m/(s^2), the vertical component will begin at 20, and decrease by 9.8 every second, or:
Vv(t) = 20 - 9.8t

Now you need to find the time at which the horizontal position (the integral of Vh(t)) equals 50, as that is the time at which the stream of water will hit the building. Once you have that time (haven't done the math, but it should be around 1.3 seconds), take the integral of Vv(t) to find the vertical position function, and plug in the value of t from the previous step.

rudi_in
Jan 18, 2007, 09:46 PM
Thank you for posting your question to the Ask Me Help Desk.

Perhaps this shall turn out exactly the same as the previous answer but here is another way to solve this problem.

1. Solve for the horizontal component

Vxo = VoCOSΘ

2. Solve for time

X = Vxot

3. Solve for the vertical component

Vyo = VoSINΘ

4. Solve for the height

y = Vyot - 1/2gtē



Let me know what answer you get. :)