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Free falling object with gravity

  1. Sep 25, 2011 #1
    I was trying to do this problem, but I got t negative, so I went back from the beginning and looked for my errors, but I couldn't find one. Can anyone tell me what I did wrong ? Thanks

    You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed Vf. You now repeat the drop, but you have friend on the street below throw another ball upward at speed Vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 28.7m. At what time do they pass each other ?

    Take upwards as positive, yo = 0 will be ground, H will be the distance when two balls pass each other.

    For the ball going downward:
    Vf = Vo - gt
    Vf = - gt

    -(28.7 - H) = -1/2 gt^2
    H = - 1/2gt^2 + 28.7 (1)

    For the ball going upwards

    H = Vf *t - 1/2gt^2
    H = - gt^2 - 1/2 gt^2 (2)

    From (1) and (2)

    -t^2 = 28.7/9.8
    --------> negative ?????
     
    Last edited: Sep 26, 2011
  2. jcsd
  3. Sep 26, 2011 #2
    More than one thing has gone wrong.

    First of all, your Vf is wrong.
    Hint: What does the symbol t mean? The time measurement of what?

    Suppose you found the correct Vf. You have to be careful when plugging it in.
    Hint: Which direction is positive?
     
  4. Sep 26, 2011 #3
    I was taking upwards as positive, and t will be the time when 2 balls pass each other. Can you tell me why the first one Vf is wrong ???

    Do you have to choose two directions ? Because at first, you have one of the ball drop, and second you have one ball drop downward and another ball throw upward ??
     
    Last edited: Sep 26, 2011
  5. Sep 26, 2011 #4
    Because t is not the time it takes for the first ball to fall to the ground.

    You have to consider the direction the ball is being thrown.
     
  6. Sep 26, 2011 #5
    But we have Vf = Vo + at, and in this case, its falling down under gravity, so a = -g ? t could be anywhere when the ball drop till it hit the grounds right ? Vf will increase as long as t increases until the ball hits the ground.

    I was taking upwards as positive
     
  7. Sep 26, 2011 #6
    But t is the time it takes for the falling ball to reach height H above ground. Are you saying the speed at that time is the same speed the ball will reach the ground?
     
  8. Sep 26, 2011 #7
    Oh, I got it, the time is different in 2 cases. Now I understand it. So, in order to find Vf, you should use
    Vf^2 - Vo^2 = 2aH with Vo = 0

    One more question, if I choose upwards as positive, it should be
    Vf^2 = -2gH or - Vf^2 = - 2gH ???

    and if downwards is positive
    Vf^2 = 2gh ?

    I'm kinda confused, if you choose the direction, would it affect only the sign of accelerator (a, g) or velocity ? or both ?
     
    Last edited: Sep 26, 2011
  9. Sep 26, 2011 #8
    Your choice of equation is good. But the way you plugged in the terms is still not correct. H is the height above the ground at which the balls pass each other, right? Is that the correct displacement to use when we consider the first ball falling to the ground from the window?

    You can find the answer to this question if you look at the definition of velocity and acceleration. Post the definitions, and post which ones you think will be affected by the choice of direction.

    Once you figure that out, you can find the answers to the following,
     
  10. Sep 26, 2011 #9
    Okie, so if H is the height above the ground at which the balls pass each other, and it's also the displacement when we first drop the first ball falling to the ground from the window. I choose y at ground = 0.

    Taking upwards as positive, then y0 = (28.7 - H).
    Vf^2 = -2g(y - y0)
    Vf^2 = -2g(-28.7 + H)

    If taking downwards as positive, y0 = -(28.7-H)
    Vf^2 = 2g(-28.7 + H)

    Am I doing right ?
     
  11. Sep 26, 2011 #10
    Your y0 is wrong in both cases. The height from the ground to the window isn't 28.7 m - H. That is the distance from the window to the point where the balls pass each other.
     
  12. Sep 26, 2011 #11
    Yes, you're right. I'm so dumb....and thanks again for your time with me even though I'm such an idiot......You would have a lot of patience I guess ;)

    so it'd be much simple than.

    If taking upwards as positive then
    Vf^2 = -2g(0-28.7)

    If taking downwards as positive than
    Vf^2 = 2g(28.7)

    So now I know how to put sign for the velocity and accelerator, but I'm a bit confused about the displacement.....if you have y0 to y and the direction that you choose is pointing at the same way, then that displacement will be positive and if its opposite then it'd be negative ???
     
    Last edited: Sep 26, 2011
  13. Sep 26, 2011 #12
    Don't think so. You've almost solved it by yourself. You just needed to be nudged in the right direction.
    But still not correct. Vf^2 cannot be negative. If it is negative, Vf will be an imaginary number(will have [itex]\sqrt{-1}[/itex]). A physical quantity cannot be an imaginary number. So there is something wrong with the way you've plugged in the terms to the relevent kinematic equation.

    Here is the full kinematic equation with all the terms,

    [itex]V_f^2 - V_i^2 = 2a(y_f - y_i)[/itex]
    where subscript i is for initial and subscript f is for final.

    Looking at your following question, I think you may have an idea what is wrong.
    The displacement increases in the positive direction.
     
  14. Sep 26, 2011 #13
    Ok, so if I take upwards as positive, then in my case, yo > y. So it would have been
    Vf^2 = -2g(-28.7)

    downwards
    Vf^2 = 2g(28.7)

    Yup, finally I understand....even though it takes me a day just to realize such a simple thing.


    Thanks again omoplata for being my great great teacher.
     
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