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Gravity And Balls Falling

  1. Dec 1, 2003 #1
    Ok guys.

    I've just come back from Thanksgiving holidays and I just can't seem to grasp this concept.

    I have an X, Y, and t from the experiment that we did in lab today. I have a Vy of 91.2cm. I now need to calculate a VyFinal ( Vyf ). From what I have gathered the equation should come out to :

    Vyf = Vy + g * t

    So, in that said, my equations for the four balls ( diferent sizes ) should come out to be :

    Vyf = 91.2cm + ( -9.80m/s2 * .50s ) ( my time for the ball to drop from 91.2cm high )

    This would leave me adding a (cm) and a (m/s). That doesn't seem right to me because the units are odd.

    I dropped each ball ( tennis, golf, nerf, and bouncy ) three times and recorded the time as t, the distance from which the ball traveled away from the dropping point as X, and Y being my constant at 91.2cm high off the ground. ( I believe this is going to become negative because the ball is traveling downward )

    I have to do calculations for Vyf and Vx.

    If X = (Vx)(t), then Vx = X/t correct? If this is true then my Vx would then be the distance from the counter to where the ball dropped divided by the amount of time it took to hit the ground.

    I think I am on the right direction, I just need some advice to get me going.

    Thanks Everyone,
  2. jcsd
  3. Dec 2, 2003 #2


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    I believe the formula you want to use would be:

    H = Vf*t - 1/2*a*t^2

    H = the height that the balls fell...AKA 92.4 cm.
    Vf = the final velocity of the balls, right before impact (what you're solving for)
    t = the time it took to fall
    a = the acceleration due to gravity, which is approximately equal to 9.81 m/s^2

    Sub all of that in and you should be able to solve for the final velocity before impact.
  4. Dec 2, 2003 #3
    Yes, On your first part of the question, adding cm's to m/s would be incrroect. The correct formula is

    final velocity = initial velocity + gt

    since the ball is at rest when you drop it, the initial velocity would be 0 m/s.

    Your 91.2 cm would be the initial height, not initial velocity. Earlier, you were incorrectly placing the 91.2 cm in as an initial velocity, which it isn't.

    The horizontal velocity would be the horizontal distance traveled/ time. For this part of the lab, you would be on the right track.
  5. Dec 2, 2003 #4
    Actually, it would be much easier just to go

    Vf = Vi + gt which is really just

    final velocity = g * time
    since Vi would be 0, as the object starts out at rest. Remember that g = -9.8 m/s^2
  6. Dec 2, 2003 #5
    ..hey Crackle, i was just wondering, whats the name of the high school you go to. I remember actually doing that same lab last year in my physics class.
  7. Dec 6, 2003 #6
    St. Charles Catholic

    And thanks to everyone who helped me as well.

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