# Semi-hard Projectile Motion Problem.

1. Oct 8, 2006

### AznBoi

Ok, I've just come across a confusing projectile motion problem. I have my own solution for it from the work that I have done so far. Please help me finish the problem for the information I have so far. Thanks a lot!:tongue:

Problem: A fireman, 50m away from the burning building, directs a stream of water from a ground level fire hose at an angle of 30 degrees above the horizontal. If the speed of the stream as it leaves the hose is 40 m/s, at what height will the stream of water strike the building?

My solution so far from the work I have done.
Ok, first I found the x and y independent velocity components of 40m/s at 30 degrees above horizontal. Vy=20m/s; Vx=34.64 m/s

I found the time to the highest point, 2.041s; therefore the time of the total flight is 4.082s.

Then I found x (total distance traveled) which is 141.4m

Then I found the highest point which is 20.41m

But the I realized that I had to find the height of when it reaches the building 50 meters away... I have all the information for the TOTAL FLIGHT. How am I suppose to find the height of when the distance is 50m?? Do I substitute 50m for X=Xo+Voxt+.5AxT^2?? Then I find the time of that distance, then the hight?? What am I doing wrong here??

At first, I wanted to find all the info of the TOTAL FLIGHT then try to come up with a coordinate for 50m, but now I realized that it would be hard to graph it. I don't know what to do, I haven't really done a lot of these problems before.

However, I understand the basic concepts of how to find the information of these kind of projectile motion problems. Thanks a lot for you help!:tongue:

2. Oct 8, 2006

Since, $$x(t)=v_{0x}t$$, you can easily find the time it takes to reach the building by plugging in the distance from the building and the x-component of the initial velocity. Further on, use that time in the equation of displacement for the y-direction to obtain the height.

3. Oct 8, 2006

### AznBoi

So is this correct?:

50m=Vxo*t
50m=34.64m/s*t

t=50m/(34.64m/s)

Then you plug the time in for y=Vyot+.5gt^2:

y=20m/s*t+.5(-9.8m/s^2)(t)^2

So that would be the height of where the water hits the building at 50m away?

4. Oct 8, 2006

Yes, it should be, after you plug in t = 50/34.64 sec.

5. Oct 8, 2006

### AznBoi

Alright, thats what I was thinking of before. I get it now, thanks for your help!! :tongue: I wasted all my time finding the total information =P

6. Oct 8, 2006

### Ahmedstein

you need to find the time when the horizontal displacement is 50, and then find the vertical displacement at that point.

I got t= 5/3

7. Oct 8, 2006

### Ahmedstein

maaan, you guys reply so fast!

8. Oct 8, 2006

### AznBoi

The time would be:

Avg. V= X/t

t=X/V

So the displacement would be 50m and the Avg. velocity or Vxo would be 34.64m/s. I already found the Vxo component in the beginning.

Therefore the time would be: 50m/34.64m/s :tongue:

Btw, the answer that I got for the height of the building that the water hits is: 18.66 meters.

18.66m makes sense because the highest point(total) would be 20.41m. The highest point's time is 2.041 seconds. For the time of 50m I got 1.4434s which would make the 18.66m be pretty accurate.

9. Oct 8, 2006

### AznBoi

Can anyone check to see if 18.66 m is the correct height? Thanks

10. Oct 8, 2006

### Ahmedstein

yeah dude, my bad.. but it's not the average velocity it's just the x component of V which is constant at each point. anyway

Y = 20*1.44 - 0.5*9.8*1.44^2
= 18.64 Hooooray