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Rocks thrown out of a window problem.

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    In a sticitly scientific experiment, a student athlete throws rocks out the window in all directions. All rocks have the same initial speed V. It turns out that all rocks' landing velocities make angles theta or greater with the horizontal. Find height h of the window above the ground.


    2. Relevant equations
    I used the position formula in the x direction and solved for time, i got (Xf-Xi)/V=time total.


    3. The attempt at a solution
    I am confused on how to use the angle theta, how does it come into play into this problem?
     
    Last edited: Nov 1, 2009
  2. jcsd
  3. Nov 1, 2009 #2
    Can I treat this like a projectile problem?
     
  4. Nov 1, 2009 #3

    tiny-tim

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    Welcome to PF!

    Welcome to PF! :smile:

    (have a theta: θ and an alpha: α :wink:)
    Yes, it is a projectile problem …

    a projectile is anything which experiences no force except gravity.

    Call the initial angle α, and the height h, and calculate the final angle, β, as a function of α and h.

    Then find the minimum value of β, and put it equal to θ. :wink:
     
  5. Nov 1, 2009 #4
    [PLAIN]http://i901.photobucket.com/albums/ac211/dktechguy112/physics131rockproblem.jpg[/PLAIN]

    I posted up an image of what I think the problem looks like.

    First off let me say thanks for the help, this forum is awesome!

    I am just a little bit confused when you say initial angle α and final angle β? where are those angles?
     
    Last edited by a moderator: Apr 24, 2017
  6. Nov 1, 2009 #5

    tiny-tim

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    They're exactly what it says on the tin … α is the initial (launch) angle for an individual rock and β is the final (landing) angle for the same rock.

    (and your diagram shouldn't show θ … it isn't necessarily the landing angle for any particular rock)
     
  7. Nov 1, 2009 #6
    [PLAIN]http://i901.photobucket.com/albums/ac211/dktechguy112/physics131rockproblemab.jpg[/PLAIN]

    I redid my drawing, did I accurately represent angles α, β, and θ?
     
    Last edited by a moderator: Apr 24, 2017
  8. Nov 1, 2009 #7

    tiny-tim

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    Hi physicsguy112! :wink:

    Yes, your α and β are fine (though you don't need θ in the drawing at all).

    Now split the initial velocity into x and y components, and use the standard constant acceleration equations to find β as a function of v and h and α. :smile:
     
  9. Nov 1, 2009 #8
    Do I need to take into account the fact that he threw multiple rocks, and some were at an angle greater then theta?
     
  10. Nov 2, 2009 #9

    tiny-tim

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    (just got up :zzz: …)

    No, just do one rock at a time. :smile:
     
  11. Nov 2, 2009 #10
    Thanks so much for the help. I attempted to solve it, and I got

    hequation.jpg

    it is supposed to be V*cos theta, not just cos theta

    I used this equation

    finalvelocityequation.jpg

    And delta y is the displacement of y, which I think is H, so I plugged everything into that equation and solved for H.

    Does this look right? I only get one shot a this, and I want to make sure I got it 100% before I turn it in.
     
  12. Nov 2, 2009 #11

    tiny-tim

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    Hi physicsguy112! :smile:

    Sorry, I have no idea what you've done. :redface:

    Yes, vfy2 = viy2 -2gh is correct, but I don't understand the rest at all.
     
  13. Nov 2, 2009 #12
    Thanks again for the help, I have uploaded my work and here is the image

    scan0001.jpg

    The key to my work being correct is the assumption that Voy=V*sinθ and that Vox=V*cosθ.
    I am not sure if those are right.

    I solved the position function for t.
    I solved the final velocity equation, and plugged in the position function solved for t.
    Then I plugged that into the velocity squared equation, and solved for H.

    I believe my answer must be in terms of θ and V, so I also need to figure out how to get the Px (x position) or t in terms of θ and V.
     
  14. Nov 3, 2009 #13

    tiny-tim

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    Hi physicsguy112! :smile:

    ok, now I see the working, and that px is the horizontal displacement, your "result" does look correct …

    but where does it get you?

    You don't know what px is, it's different for different initial angles, and adding an extra unknown (you should be eliminating unknowns) doesn't get you any nearer finding the landing angle.

    Hint: decide what the landing angle is, in terms of v H and the initial angle, and then use your original equation (vfy2 = viy2 -2gh) to find it.

    (oh, and you don't need any distances :wink:)
     
  15. Nov 3, 2009 #14
    so all I need to do is put Px in terms of V and θ?

    Is the Voy=V*sinθ and Vox=V*cosθ correct?

    When u say, decide what the landing angle is, do u mean pick a value for it?
     
  16. Nov 3, 2009 #15

    tiny-tim

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    No, find an equation for the (tangent of the) landing angle, involving only v H and the initial angle.
     
  17. Nov 4, 2009 #16
    I am still confused on this problem.
    Do I need to scrap my whole H= equation?

    what do you mean by tangent of the landing angle? How does it relate to the initial angle?
    My answer has to be in terms of θ, and V, so how can I use initial and final angles?
     
  18. Nov 4, 2009 #17

    tiny-tim

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    Your vfy2 = viy2 -2gh is correct.

    The tangent of the landing angle is vfy/vfx.

    You must find how this depends on the initial angle, and then find its minimum … that is the only way of finding θ (because θ is the minimum).
     
  19. Nov 4, 2009 #18
    Thank you so much for the help so far.
    I am really trying to get this one, but I am still lost.

    I realize now that
    Vox=V*sina where a=initial angle
    voy=V*cosa
    Vfx=V*sina
    Vfy=V*cosa+Ay*t
    θ=tan^-1(Vfy/Vfx)

    but I'm still confused on how to tie a and b together to get an answer in terms of θ.
     
  20. Nov 4, 2009 #19

    tiny-tim

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    Why are you persisting in introducing an extra variable (t) which isn't given?

    You have a perfectly good equation for vfy in terms of v a g and h.
    (try using the X2 tag just above the Reply box :wink:)

    No, tan-1(Vfy/Vfx) is the landing angle for initial angle a.

    θ is the minimum value of that, over all values of a.
     
  21. Nov 5, 2009 #20
    I figured it out!!!
    The trick was that I need to find the minimum angle for theta, this occurs when the initial angle =0.
    I got h=((V*tanθ)^2)/2g

    Thanks so much for your help!
     
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