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Homework Help: Semi-hard Projectile Motion Problem.

  1. Oct 8, 2006 #1
    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. jcsd
  3. Oct 8, 2006 #2


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    Homework Helper

    Since, [tex]x(t)=v_{0x}t[/tex], 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.
  4. Oct 8, 2006 #3
    So is this correct?:



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


    So that would be the height of where the water hits the building at 50m away?
  5. Oct 8, 2006 #4


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    Yes, it should be, after you plug in t = 50/34.64 sec.
  6. Oct 8, 2006 #5
    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
  7. Oct 8, 2006 #6
    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
  8. Oct 8, 2006 #7
    maaan, you guys reply so fast!
  9. Oct 8, 2006 #8
    The time would be:

    Avg. V= X/t


    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.
  10. Oct 8, 2006 #9
    Can anyone check to see if 18.66 m is the correct height? Thanks :smile:
  11. Oct 8, 2006 #10
    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
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