1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    radou

    User Avatar
    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?:

    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?
     
  5. Oct 8, 2006 #4

    radou

    User Avatar
    Homework Helper

    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

    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.
     
  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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Semi-hard Projectile Motion Problem.
  1. Projectile Motion HARD (Replies: 16)

Loading...