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Book toss Kinematics motion

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data


    A book is toss into your dorm room, just clearing a windowsill 4.2m above ground.
    a)If the book leaves your hand 1.5m above ground, how fast must it be going to clear the sill?
    b)How long after it leaves your hand will it hit the floor, 0.87m below the windowsill?



    2. Relevant equations

    none.

    3. The attempt at a solution

    a)vf^2 - vi^2 = 2g(yf - yi)

    -vi^2 = 2(-9.8ms^-2) (4.2m-1.5m)
    vi = 7.274613392ms^-1


    b)vi = 7.274613392ms^-1
    yf - yi = 4.2m - 1.5m (are the directions correct?)
    g = -9.81ms^-2

    yf - yi = vit + 0.5gt^2

    3.33m - 1.5m = 7.274613392ms^-1 t - 4.905 ms^-2 t^2

    solving via quadratic equation:

    t = 1.162s && t = 0.32

    Which t do I take? Conceptual explanation please?
     
    Last edited: Dec 15, 2013
  2. jcsd
  3. Dec 15, 2013 #2

    Curious3141

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    The later one (1.162s). The earlier t signifies the time the book passes the spot 3.33m above the floor on its upward trajectory (on the way to the windowsill). The later t is the time the book hits the floor outside.
     
  4. Dec 15, 2013 #3
    And I presume the directions of yf and yi in part b is both correct?

    I would have made the assumption that at yi = 1.5m and going up, the displacement is positive and at yf = 3.33m, the displacement would be negative.
     
  5. Dec 15, 2013 #4

    Curious3141

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    The important thing for the displacement is that you define an origin point (in this case, it's the floor directly below your hand) and a positive direction (in this case "up").

    So the initial displacement of the book is +1.5m. The final displacement is +3.33m (because it ends up 3.33m above the level of the dorm room ground).

    The problem is also completely workable if you take the origin as your hand. In which case the original displacement is 0, and the final displacement is +1.83m. The only thing that matters is consistency.
     
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