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Homework Help: Ball into cup off ramp, need to derive equation

  1. Feb 5, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem is to find at what distance the ball has to be placed at on a ramp given a certain angle measurement of said ramp. The cup's position will be given a few minutes prior to testing so there are more unknown variables.
    Table Height: .73M
    Cup Height: 10cm
    Angle measurement is 40 degrees
    Ball Mass: 3.6 grams
    .3: rotational, so we are supposed to only use 70% of the energy calculated
    We are required to derive an equation to use at the spot to figure out how far up the ball needs to be (given the cup distance)

    2. Relevant equations
    X=(1/2) A*T^2+ViT+Xi
    V= sq root (2gh)

    3. The attempt at a solution
    Ive used the X=... equation and entered the square root equation into it, but i still cannot figure out how to derive it. Ive figured out the sin/cos which is .6428 and .7660 (I would just multiply then to get distance).
    Any hints would be greatly appreciated. If i've left anything out or forgot to mention some key element, please state that.
    Last edited: Feb 5, 2010
  2. jcsd
  3. Feb 5, 2010 #2


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    Homework Helper

    Um, what is the question, exactly? What you've written is very confusing.
  4. Feb 6, 2010 #3
    I apologize for not being concise and coherent. But I need to derive an equation so that when I put in the location of the cup, I can figure out how far up the ball should go on the ramp. The ramp is 40 degrees, height of table is .73m, height of the cup is 10 cm or .1m, and the mass of the ball is 3.6grams. We are only supposed to use 70% of the energy calculated because 30% is lost when the ball is moving/rolling. I can show some more of my work but I haven't progressed, I continue to get an identity...which is becoming very frustrating. I dont know If i am explaining it any better.
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