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Projectile Motion Water hose Question

  1. Oct 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D . The hose shoots the water at 45 ∘ above the horizontal from the same level as the base of the tank and is a distance 6D away


    2. Relevant equations
    b1c5647476c66437d2183de53f091d77.png
    8f817acb7ed26dd699a14912a0125f23.png
    Vx=Vt
    D=vit+1/2at^2

    3. The attempt at a solution
    I tried plugging in the values into the 8f817acb7ed26dd699a14912a0125f23.png equation and I end up with√6dG<V<√√7dG
     
  2. jcsd
  3. Oct 1, 2015 #2

    haruspex

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    You should not apply equations without understanding the context in which they are valid. The equation you tried does not allow you to use the height of the tank, so it cannot be appropriate. When is it appropriate?
     
  4. Oct 1, 2015 #3
    Right, okay I think I'm only allowed to use the equation when the object is landing at the same level ground as it started off on.
     
  5. Oct 1, 2015 #4

    haruspex

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    Right. Or, more precisely, when d is the distance to a point where it is at the same level it started at.
     
  6. Oct 1, 2015 #5

    gneill

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    What is the question? What are you trying to find?
     
  7. Oct 1, 2015 #6
     
  8. Oct 1, 2015 #7
    oops. Sorry I thought I posted that.
     
  9. Oct 1, 2015 #8

    gneill

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    Okay, so what are the scenarios for the water stream that you need to investigate? That is, what are the significant characteristics of the trajectory in each case?
     
  10. Oct 1, 2015 #9
    OK. From where is the hose 6D away? From the centre of the tank? Or from where the tank begins?
     
  11. Oct 1, 2015 #10

    gneill

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    Clever student's strategy: Unless the problem states a condition explicitly, choose the simplest situation to work with and declare it as an assumption. Just make sure that it's really undefined and not subtly implied somehow in the problem statement.
     
  12. Oct 1, 2015 #11
    You can't use dg = v^2(sin∆) here as this is equation gives us the maximum range of the projectile.
     
  13. Oct 1, 2015 #12

    haruspex

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    See post #3.
     
  14. Oct 1, 2015 #13

    haruspex

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    So what equation should you use here?
     
  15. Oct 1, 2015 #14
    I used the 2D equations and plugged in t=Vx/Vcos45 into d=vtsin45+1/2at^2 and solved.

    Thanks
     
  16. Oct 1, 2015 #15

    haruspex

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    So you got the answer?
     
  17. Oct 1, 2015 #16
    yep
     
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