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Homework Help: Physics 12 Projectiles Question

  1. Jul 12, 2014 #1
    1. The problem statement, all variables and given/known data

    A marble is accelerated to a horizontal velocity of 0.50 m/s by rolling it down a small ramp. The marble rolls off the table, which is 0.78 m high. As it falls, it hits a barrier on the way down. If the barrier is 0.15 m from the edge of the table, at what height from the ground, h, will the marble hit the barrier?

    2. Relevant equations

    d = (Vi)(t) + 1/2(a)t/2

    Vf =Vi + at

    (Vf)2 = (Vi)2 + 2ad

    3. The attempt at a solution

    I wasn't really sure how to start it, but first I found the time it would take for the marble to fall off the table:

    Using vertical components:

    d = (Vi)(t) + 1/2(a)(t)2
    -0.78 = (0)(t) + (1/2)(-9.8)t2
    -0.78 = -4.9t2
    t = 0.40s

    I don't know if that ^ is right, and I'm not really sure what I should do next. Any help would be greatly appreciated!
  2. jcsd
  3. Jul 12, 2014 #2


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    You know how far the barrier is from the edge of the table. If the marble is traveling 0.5 m/s horizontally when it leaves the edge, how long will it take the marble to cross the gap between the edge and the barrier? How far does the marble fall in this amount of time?
  4. Jul 12, 2014 #3


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    That would be the time taken to hit the floor not the barrier.

    Have a look at what information the problem gives you about the horizontal components.

    (cross posted with steamking)
  5. Jul 12, 2014 #4
    I'll take downward as my positive direction. But first, I calculate the time it takes to reach the barrier: 0.15/0.5=0.3. Then my initial vertical velocity is zero, and the time of fall would be 0.3 s. I will take acceleration of free fall to be 9.81 m s^-2. Then displacement from top of the table = (initial velocity)(time of travel) + (0.5)(acceleration)(time)^2. That in symbols would be s=ut+0.5a(t^2). Then s = (0)(0.3)+(0.5)(9.81)(0.3^2). This comes out to be 0.44 m from the top of the table. Then height h would be 0.78-0.44 that would be 0.34. I hope that is right. =P
  6. Jul 13, 2014 #5


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    Should let the OP work it out.
  7. Jul 13, 2014 #6
    Oh. I didn't know it worked like that. Sorry. I'll take care now.
  8. Jul 14, 2014 #7
    Thanks everyone! This problem definitely wasn't as difficult once I properly identified vertical and horizontal components.
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