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Ball Thrown up in air, Velocity Question

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown vertically upward with an initial speed of 11 m/s. One second
    later, a stone is thrown vertically upward with an initial speed of 25 m/s. (a) Find
    the time it takes the stone to catch up with the ball. (b) Find the velocities of the
    stone and the ball when they are at the same height.

    2. Relevant equations



    3. The attempt at a solution

    a = 9.8
    Vi,ball= 11 m/s Vi,stone= 25m/s

    y= y0+ V0t+ 1/2 at2

    y = 11t+ 1/2 (9.8) t2

    y= 11t - 4.9 t2 ======= ball

    y = 25 (t+1) + (4.9 (t+1))2

    y = 25t + 25 - 4.9 ( t2 + 2t + 1)

    y = 25 t + 25 - 4.9 t2-9.8t - 4.9

    y= -4.9t2 + 15.2t + 20.1 ===== stone

    11t - 4.9 t2 = -4.9t2 + 15.2t + 20.1



    am i on the right track??
     
  2. jcsd
  3. Dec 11, 2008 #2

    LowlyPion

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    Sort of.

    I think it might be easier to take a snap shot at the start of the second launching.

    At that moment the ball is

    Yo = 11*t - ½gt² evaluated at 1 sec

    And at 1 sec, the velocity V1 = 11 - 9.8*t evaluated at 1 sec.

    Now you can write your equations all with the same time can't you?

    The ball now can be given by:

    Y = (11 - ½g) + (11-9.8)*t - ½gt²

    The rock:

    Y = 0 + 25*t - ½gt²

    When they are equal:

    25*t - ½gt² = (11-4.9) + 1.2*t - ½gt²

    And it looks like the ½gt² disappears.
     
  4. Dec 12, 2008 #3
    so i get t=0.256 .... is this right? so this is the time it takes for the stone to catch up to the ball. how could i figure out B) the velocity?
     
  5. Dec 12, 2008 #4

    turin

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    You can check your answer for reasonability by drawing a graph. Do you know what kind of curve these trajectories make in the y,t plane? Draw both of them on the same graph (with the proper shift), and you should see a particular point of interest.
     
  6. Dec 12, 2008 #5
    sorry, i have absolutely no idea how to draw that lol
     
  7. Dec 12, 2008 #6

    turin

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    Hmm. What do you mean by "this"? Of course, the first step would be to decide what is the shape. A mathematical equation y=f(t) for each of the two balls would suffice. Then, you can either use the (antiquated, but useful on tests) skill of sketching this out by hand, or you can use Excel (or some other spreadsheet/plotting program).
     
  8. Dec 12, 2008 #7

    LowlyPion

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    What is the equation for the rock velocity?

    V = Vo - a*t

    You know t, so ...
     
  9. Dec 14, 2008 #8
    should i add 1 second to my time?
     
  10. Dec 14, 2008 #9

    LowlyPion

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    You would add 1 sec if you wanted to know how long the ball was in the air.
     
  11. Dec 14, 2008 #10
    o ok.. i think i understand now.. so my answer is then i guess t= 0.256, since it was thrown 1 second after.
     
  12. Dec 14, 2008 #11

    LowlyPion

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    This was originally why I suggested you consider taking the snapshot at the moment the rock was thrown and solving for t, because that's the answer you are looking for.
     
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