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Free Fall Trajectory? Simple answer!

  1. Nov 6, 2008 #1
    Free Fall Trajectory

    So I am basically lost in what that means. I think it is the component method for vectors? Am I right? Some background on the lab: We rolled a marble down a curved ramp so that it went off the table and hit the ground in front of the table. (we marked where it hit with carbon copy paper. I already measured the potential energy and velocity.

    1. The problem statement, all variables and given/known data
    Question: Recalculate the horizontal marble velocity of the marble at the bottom of the track using the free fall trajectory method.

    The height of the curved track from the table was 14.3cm and the height from the table to the ground was 84.4cm
    And the distance that the marble traveled from the edge of the desk was 44.5cm
    mass of marble = 13.7g

    Do I need the length of the track?

    2. Relevant equations
    (THis is what I am looking for)

    3. The attempt at a solution
    If someone could explain what free fall trajectory method is (my teacher doesn't teach, he just talks). That should be enough to point me in the right direction. Also if it is the component method, then i would need the angle of the ramp, correct?

    Last edited: Nov 6, 2008
  2. jcsd
  3. Nov 6, 2008 #2


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    Homework Helper

    You basically have it, so long as you know the height of the table.

    Knowing the height you can calculate the time that it takes to drop.

    Your Vy is initially zero until it goes off the edge.

    So the time is given simply by 1/*g*t2 = height of table.

    Knowing the time you're rolling in clover because the x distance isd calculated simply by Vx * time.

    Not asked but going further if you know the V you know the m*v2/2 which is going to be equal (if you could ignore friction and such) to the initial Potential energy of the marble at the height it started rolling down from.
  4. Nov 6, 2008 #3
    Ok, so when I do:
    height of table = 84.4cm


    I get t=.415s

    Then; dx=Vx*t gets me Vx of 1.07m/s

    Thats the velocity in the X-direction?

    If so then my other calculations are off by 400%....

    This is what I got for my other calcs:
    htable and ramp = 98.7cm

    PE=(.0137kg)(9.80m/s)(.987m) =.133 J

    .133 J=.5*m*v^2
    .133 J= .5*(.0137kg)*v^2
    v=4.41 m/s

    Is all of that right?

  5. Nov 6, 2008 #4


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    Homework Helper

    Not quite. If

    1/2*m*v2 = m*g*h

    Then your initial Vx will be given by Vx = (2*g*h)1/2

    The h though is only the h of the ramp down to the top of the table.

    That comes to (2*9.8*.143)1/2
  6. Nov 7, 2008 #5
    Oh. Alrighty. I guess it just clicked in my head. Thanks.
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