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Basic kinematics: projectile motion

  1. Nov 11, 2015 #1
    This is a really basic problem but the answers I've gotten don't match up with the back of the book. I've been over it several times & can't understand where I'm going wrong - any pointers would be much appreciated!

    1. The problem statement, all variables and given/known data

    A physics book slides off a horizontal tabletop with a speed of 1.40m/s. It strikes the floor in 0.320s. Ignore air resistance. Find a) the height of the tabletop off the floor; b) the horizontal distance from the edge of the table to the point where the book strikes the floor.

    2. Relevant equations
    vf = vi + at
    x = xi + vit + ½at2

    3. The attempt at a solution
    I broke up the book's initial velocity into components:
    Vf-x = Vi-x = 1.40m/s
    Vf-y = Vi-y +at = -gt = -3.14m/s

    The height of the table = the magnitude of the vertical displacement y (using the top of the table as the origin):
    y = yi + Vi-yt + ½at2 = 0 + 0 + ½(-9.8m/s2)(0.320s)2= -0.502m -> height = 0.502m
    What the book gives for this is 1.13m.

    The horizontal distance = magnitude of horizontal displacement x:
    x = xi + Vi-xt + ½at2 = 0 + (1.40m/s)(0.320s) + 0 = 0.448m
    What the book gives here is 0.502m (which is what I got for the height - which means I'm either completely mixed up or maybe the book is wrong?!)

    Thanks for taking a look.
     
    Last edited: Nov 11, 2015
  2. jcsd
  3. Nov 11, 2015 #2

    haruspex

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    How did you get vfy=-4.9m/s? Looks like you used t=.5s, which, as it happens, would give a height closer to 1.1m.
     
  4. Nov 11, 2015 #3
    Yes, you're right, sorry! I've been trying other problems since getting stuck on this one & copied the wrong figure down from my rough work. It should be Vf-y = -3.136m/s

    I'll edit that now.
     
  5. Nov 11, 2015 #4

    haruspex

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    Then I agree with your 0.5m height.
     
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