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Kinematics in two dimensions problem

  1. Oct 9, 2009 #1
    A fire fighter directs a stream of water from a fire hose into the window of a burning building. The window is 20.0 m above the level of the nozzle and 60.0 m away. The hose produces a jet of water at a constant nozzle speed of v0 = 30.0 m/s. The fire fighter can control the angle θ0 that the water is aimed at by moving the nozzle with her hands.

    (a) The fire fighter discovers that there are exactly two angles θ0 that she can hold the hose for which the water gets into the window. What are the two angles?

    (b) The fire fighter further discovers that if she backs slowly away from the building there is a distance Dmax where there is only one angle θm which gets the water into the window. If she goes any farther back, she cannot get the water into the window at all. Find Dmax and θm.
     
  2. jcsd
  3. Oct 9, 2009 #2

    tiny-tim

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    Hi Dmmd! Welcome to PF! :wink:

    Use the standard constant acceleration equations in the x and y directions separately …

    what do you get? :smile:
     
  4. Oct 9, 2009 #3
    You mean Vox and Voy? after i take the components of velocity? This is where I'm stuck since there are 2 variables that I don't have which is the time and the angle
     
  5. Oct 10, 2009 #4

    tiny-tim

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    ok, call the time t and angle θ …

    eliminate t from the two equations (it will be the same t, since obviously the water arrives horizontally and vertically at the same time! :wink:), and that should give you an equation with only 1 variable (θ). :smile:
     
  6. Oct 10, 2009 #5
    although you're right on the statement that the time is the same for x and y i can't cancel the time in the equation since the variance of time is tf-ti. Hence ti=0 but tf is different from 0. So this results in 2 other equations which i'm not sure if they're correct. so what i came up is that i should use x=vox.t and y=voy.t+1/2g.t^2. Is this correct?
     
  7. Oct 10, 2009 #6

    tiny-tim

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    Yes, and now put the numbers (and the cosθ and sinθ) in, and then eliminate t. :smile:
     
  8. Oct 11, 2009 #7
    But after eliminating t. I couldn't solve the trig orz....
    I tried so many trig identities.
    the equation looks like this
    cos^2(x)=3sinxcosx-0.98
     
  9. Oct 11, 2009 #8
    Ok I solved the first part....
    However,
    I couldn't find a way to solve the second part.
    Anyone can give me a head start?
     
  10. Oct 12, 2009 #9

    tiny-tim

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    Hi hdlineage! :smile:

    (just go up :zzz: …)

    In your equation with cos2θ etc, instead of 60, write the horizontal distance as x.

    Then you're looking for the value of x which gives that equation two identical roots. :wink:
     
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