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Intro Projectile Motion

  1. Sep 24, 2007 #1
    Having some trouble with the last part of this question:

    A ball is projected horizontally from the edge of a table that is 1.11 m high, and it strikes the floor at a point 1.36 m from the base of the table. What is the initial speed of the ball?

    How high is the ball above the floor when its velocity vector makes a 46.4o angle with the horizontal?


    So I've got the answer to the first part (yea me) by doing this:

    First I solved for time, and got an answer of 0.476 seconds (I know I'm cheating by not typing the wquation out but I'm not sure how to type out square roots and intiial velocities and all those lovely little signs)

    Then I went:

    Delta x = v(intitial)(cos theda int)t
    v=1.36 m / 0.476 s
    v = 2.86 m/s

    So I've got my initial velocity of 2.86 m/s.


    So my problem comes with this second part of the question:How high is the ball above the floor when its velocity vector makes a 46.4o angle with the horizontal?

    So I think I would be looking for Vy at 46.4 degrees, but I'm not sure how to find that, because all of the equations I know only contain initial theda, not some random theda along the way. Is this solved witha projectile motion equation, is it a purely vector related question, or a combination of both? Should I be using other information in the question?

    Any guidance is greatly appreciated!

    Thanks!
     
  2. jcsd
  3. Sep 24, 2007 #2

    Doc Al

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    Staff: Mentor

    Hint: What value of Vy will give you the required angle?
     
  4. Sep 24, 2007 #3
    hmmm I need more of a hint I'm afriad...I'm not sure how I can find Vy if I only have the angle...I don't know what the size of the vector is (do I?), or otherwise I could go
    Vy=Vsin(theda)
     
  5. Sep 24, 2007 #4

    learningphysics

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    If you know the angle is 46.4... what do you know about tan(46.4) ? or rather tan(-46.4)
     
  6. Sep 25, 2007 #5
    hmmm I know that tan(-46.4)=(Vy / Vx) So if I calculate that out I get tan -46.4 = -1.05, but I'm not sure what that means. Is that the Vy value, or do I still have steps to go?
     
  7. Sep 25, 2007 #6

    Doc Al

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    You have another step to go. The tangent equals the ratio Vy/Vx. You found the tangent, and you know Vx, so you can solve for Vy.
     
  8. Sep 25, 2007 #7
    Ok this may sound really lame (don't worry, I'm used to it lol) but how do I know what Vx is? The only x value I know is the total change in distance along the x-axis, which is 1.36 m. That's not the Vx value is it? Or is the velocity that I found (2.86 m/s) Vx?
     
  9. Sep 25, 2007 #8

    Doc Al

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    Yes. That's the horizontal component of the velocity, which remains constant (gravity affects only the vertical component).
     
  10. Sep 25, 2007 #9
    so if tan(-46.4)=(Vy / Vx) ;
    -1.05 (2.86 m/s) = Vy
    -3.00 m/s = Vy ?
     
  11. Sep 25, 2007 #10

    Doc Al

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    Looks good to me.
     
  12. Sep 25, 2007 #11
    that's good :D But I'm still confused about how to answer the question.

    How high is the ball above the floor when its velocity vector makes a 46.4o angle with the horizontal?
     
  13. Sep 25, 2007 #12

    Doc Al

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    Answer this question instead: Where is the ball when its vertical speed is -3.00 m/s?
     
  14. Sep 25, 2007 #13
    I don't know how to answer that question :( Does it use a projectile motion equation?
     
  15. Sep 25, 2007 #14

    Doc Al

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    Yes. You'll need the kinematic equation describing accelerated motion. The vertical component of the projectile's motion is accelerated.
     
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