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Throwing a Ball

  1. Feb 10, 2013 #1

    VU2

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    Q1) You throw a ball. Assume that the origin is on the ground, with +y-axis pointing upward. Just after the ball leaves your hand its positon is <.06,1.03,0>m. The average velocity of the ball over the next .7s is <17,4,6>m/s. At time .7s after the ball leaves your hand, what is the height of the ball above the ground.

    y=1.03+4(.7)-9.8/2×(.7^2)
    y=1.429 is my answer, is this correct? I'm using the average velocity as initial, so i'm not sure if I can assume that or not.
     
    Last edited: Feb 11, 2013
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  3. Feb 10, 2013 #2

    Simon Bridge

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    Welcome to PF;
    Probably not.
    Check it to see if it matters. Depends how accurate your answer needs to be.
    You know the acceleration and the average velocity, can you find the y-component of the initial instantaneous velocity.
    Note: how is "average velocity" calculated?
     
    Last edited: Feb 10, 2013
  4. Feb 11, 2013 #3

    VU2

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    AvgV is calculated by (finalV+initalV)/2=AvgV, can i assume that initial V is 0?
     
  5. Feb 11, 2013 #4

    tms

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    No, you can't assume that. You can, however, calculate the final velocity in terms of the initial velocity, given the acceleration and the time. That will allow you to get the initial velocity from the average.
     
  6. Feb 11, 2013 #5

    VU2

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    Oh I see, y-final=y-initial + v(avg)*(t). Therefore, y-final=1.03+4(.7)=3.83?
     
  7. Feb 11, 2013 #6

    tms

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    That will work, too, more directly than what I suggested. I don't know what the notation "<17,4,6>" means, so I don't know if your answer is right.
     
  8. Feb 11, 2013 #7

    VU2

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    That is the average velocity vector.
     
  9. Feb 11, 2013 #8

    VU2

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    I'm curious now, how would you compute it your way?
     
  10. Feb 11, 2013 #9

    tms

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    That's what I thought, but I was wondering why the z direction was there, since it isn't mentioned in the rest of the problem.
     
  11. Feb 11, 2013 #10

    tms

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    I explained it above. It is essentially the same, except that it takes a roundabout way to get there (basically deriving the equation you used).
     
  12. Feb 11, 2013 #11

    VU2

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    Oh I see, thanks!
     
  13. Feb 11, 2013 #12

    Simon Bridge

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    Average velocity is change in position over change in time:
    ##\bar{v}=\Delta y/\Delta t = (y_f-y_i)/\Delta t##
    ... you rearranged that equation to give you the final height given the initial height and the average velocity ... well done.

    As an exercise - how would it have been different for the same figures, except the average velocity was timed over 0.1s, but you still want the final height after 0.7s?
     
  14. Feb 14, 2013 #13

    VU2

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    Simon Bridge,

    Sorry, I was suppose to get back to you sooner. That's an interesting question. I'm not sure how I would approach it, to be honest.
     
  15. Feb 15, 2013 #14

    Simon Bridge

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    Well - in that case you'd have two time periods ot consider: one short one (with a displacement) to establish the initial velocity and another, longer, one to find the answer. Otherwise the approach is identical to the one you used above.

    IRL you often measure the speed of something by timing it over a short distance, and then use that information to work out where the thing will end up.
     
  16. Feb 15, 2013 #15

    VU2

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    Simon Bridge,

    Yeah that's what I figured too. Thanks for the lesson.
     
  17. Feb 15, 2013 #16

    Simon Bridge

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    No worries - have fun :D
     
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