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Height at which objects will collide

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data

    A rubber ball is shot straight up from the ground with speed Vo. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest.

    1) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of Vo, h, and g.

    2) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?

    3) For what value of h does the collision occur at the instant when the first ball is at its highest point?


    2. Relevant equations

    Vf^2 = Vi^2 + 2a(h)


    3. The attempt at a solution

    I am not sure how to make make the symbolic expression

    Ball 1:

    Vf^2 = Vi^2 + 2a(h)
    h= (Vf^2 - Vo^2) / 2a
    h= (Vf^2 - Vo^2) / 2(-9.8m/s)
    Ball 2:

    Vf^2 = Vi^2 + 2a(h)
    h= (Vf^2 - Vo^2) / 2a
    h= (Vf^2 - 0) / 2(9.8m/s)

    Then make H=H

    (Vf^2 - Vo^2) / 2(-9.8m/s) = (Vf^2 - 0) / 2(9.8m/s)
     
  2. jcsd
  3. Sep 22, 2008 #2

    tiny-tim

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    Hi 05holtel! :smile:

    This can't possibly work :frown:

    Hint: you need an equation for h that doesn't involve Vf. :smile:
     
  4. Sep 22, 2008 #3
    Hey Tiny Tim,

    OK So I tried the formula
    sf=si +vi(t2-t1) + 1/2(a)(t2-t1)^2

    I get:

    Ball 1:
    sf = 0 +v0(delta t) + 0.5(-9.8)(delta T)^2
    = vo(delta t) - 4.9(delta t)^2

    Ball 2
    sf = h + (0)(delta t) + 0.5(9.8)(delta t)^2
    = h + 4.9(delta t)^2

    vo(delta t) - 4.9(delta t)^2 = h + 4.9(delta t)^2

    I am not sure what to do because the answer should be in a symbolic expression in terms of Vo, h, and g.
     
  5. Sep 22, 2008 #4

    cepheid

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    Try this. Let y1 represent the height of the first ball, and y2 the height of the second ball. (We are using the letter y because that is the direction in our coordinate system along which all the movement takes place).

    at ANY time t,

    [tex] y_1 = v_{0}t - \frac{1}{2}gt^2 [/tex]

    because the first ball initially starts off at yi1 = 0 and is thrown up with velocity v0, but begins decelerating immediately due to gravity.

    Similarly, at ANY time t, the position of the second ball is given by:

    [tex] y_2 = h - \frac{1}{2}gt^2 [/tex]

    because the ball starts out at yi2 = h and is just dropped, with NO initial velocity. It immediately begins accelerating downward due to gravity.

    At the collision point, clearly the positions of the two balls are the same:

    y1 = y2

    Solve for the time t at which this occurs, and then plug that time t into either formula to find the height at which it occurs.
     
  6. Sep 23, 2008 #5

    tiny-tim

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    Hi 05holtel! :smile:

    (have a delta: ∆ and a squared: ² :smile:)

    Yes, that's perfect … you got the same equations as cepheid :smile:

    except that you wrote 9.8 instead of g (which you shouldn't have done, since the question asks you to keep g in the asnwer :wink:)

    You have: vo(∆t) - g(∆t)² = h + g(∆t)² …

    just move it around until you have a quadratic equation in ∆t,

    and then solve for ∆t. :smile:

    ("symbolic" just means "no numbers")

    (btw, you could have used t instead of ∆t, like cepheid, so long as you specified that the initial time is 0 :wink:)
     
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