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Help with free falling object problem

  1. Sep 22, 2007 #1
    I hope this is the right forum, so I apologize in advance if I'm mistaken.

    Here's the question poised:

    How many seconds apart should two balls be released so that a ball being dropped from a height of H reaches the ground at the same time as a ball dropped from a height of 5H?

    I need some help on this one. Thank you.
     
  2. jcsd
  3. Sep 22, 2007 #2
    Uhhh, what kind of drag are we supposed to assume? Are you supposed to give one an initial velocity?

    I don't think this is possible even with drag, the ball from H would need lift too. Unless H is something like 1000m and the drag is nonlinear from a really massive object, would that even do it...?
     
  4. Sep 22, 2007 #3
    Welcome to the forums.

    First thing you must show your attempt. Whatever it may be.

    Now for the question, are you familiar with the equations of kinematics.
     
  5. Sep 22, 2007 #4
    Its certainly physically possible, but he's probably missing some info.
     
  6. Sep 23, 2007 #5
    nope, that's the question. I'm fairly sure it involves finding a ratio, due to the many unknowns, so I imagine finding the value of H is not the goal. It's also assumed to be no air resistance. The two balls are released or dropped, so Vo = 0 m/s for both balls.

    Show some work? OK.

    Vi1 = Vi2 = 0 m/s

    Xi1 = H; Xi2 = 5H; Xf1 = Xf2 = 0 m

    ti1 = ti2 = 0 s

    a = -9.8 m/s^2 = g

    tf1 = ? ; tf2 = ? ; Vf1 = ? ; Vf2 = ? ; tf2-tf1 = ?

    ...

    If I knew H, I could solve for tf1, tf2, and find the difference of the two

    O --> ball with Xi2 = 5H
    | |
    | |
    | | --> With H, I can find this difference: tf2 - tf1
    | |
    | |
    | O --> ball with Xi1 = H
    |
    |
    |
    ---------------------- ground, 0 meters


    Xf1 = Xi1 + Vi1*ti1 + 0.5g*tf1^2

    0 = H + 0 + 0.5g*tf1^2 ==> H = -0.5g*tf1^2 ==> tf1 = sqrt(-2H/g)

    Xf2 = Xi2 + Vi2*ti2 + 0.5g*tf2^2

    0 = 5H + 0 + 0.5g*tf1^2 ==> H = -10g*tf2^2 ==> tf2 = sqrt(-10H/g)

    tf2-tf1 = ?

    Finding the ratio of tf2/tf1 makes no sense and I know it's 5. If i square tf2-tf1, I'm changing the expression. So, at this point, I'm a bit stuck...
     
  7. Sep 23, 2007 #6
    does anybody have any suggestions or anything at all?
     
  8. Sep 23, 2007 #7
    now here is the equation of free fall

    [tex] X(t) = X0 - V0 * t - \frac{g*t^2}{2}[/tex]

    apply it to the two ball, remember that you have a delay between the two droppings.

    -----------------------------------------------------
    Correct me if I am wrong.
    http://ghazi.bousselmi.googlepages.com/présentation2[/QUOTE]
     
  9. Sep 23, 2007 #8
    [/QUOTE]

    Um, did you read my post at all? I know this formula and I know the difference between the two times is what I'm looking for. The problem is I'm stuck because the fact that I don't have a specific value for the height complicates things.

    I know that tf2 = sqrt(5)*tf1. But I can't figure out a way to use this to my advantage.
     
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