1. Oct 3, 2006

### studentmom

The problem states:

A fireman is standing 50.0 m away from a burning building. He directs a stream of water from the fire hose at an angle of 30 degrees above the horizontal. If the initial speed is 40.0 m/s, at what height will the stream of water strike the building?

I understand that the velocity (40.0 m/s) needs to be broken down into its x- and y- components using sin and cos. I imagine that I need to find a time (in seconds) in order to get the vertical height. But I can't seem to figure out the equations to use and when to use them. The x-component of the velocity is 34.6 and the y-component is 20. I know that horizontal acceleration is zero and the vertical acceleration is -g.

Thanks!

2. Oct 3, 2006

### Saketh

You want to find how much time it takes for the water to reach the wall in the x-direction first.
Then, you want to find how high the water goes in that amount of time.

So first, we want to find how much time the water needs to reach the wall in the x-direction. The fireman is standing 50m away from the burning building - now all we need is a velocity in the x-direction to find how long it takes for the water to reach the wall. Hmm. How can we do that?

After that, it's just a matter of finding how high up the wall the water goes. For this, remember that acceleration time-integrated twice gives you $$\Delta x = v_0t + \frac{1}{2}at^2$$.

Don't think about in terms of formulas at first. Think about "What variables do I need to solve for?" and then the formulas will come naturally.

3. Oct 3, 2006

### studentmom

Thank you for your help... I knew what I needed to get, but I couldn't remember how to get there... I was having trouble with the easiest part, v=displacement/time...

Once again, thank you for your assistance.