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Stone launched upward - Kinematics question

  1. Jan 27, 2017 #1
    1. The problem statement, all variables and given/known data

    Laura propels a stone straight up with her slingshot. The stone strikes a tree branch 4.0m above her. It strikes the branch with a velocity of 5.0m/s. How long does this take?

    a= -9.81 m/s2
    v1 = 5.0m/s
    d= 4.0 m


    2. Relevant equations
    d=vit+1/2at2

    d=v1+v2/2 x t
    v2=v1+at


    3. The attempt at a solution
    ATTEMPT 1: DOESN'T WORK
    I believe I should use d=vit+1/2at2
    therefore:
    4m = (5.0)(t)+1/2(-9.81)(t)2
    -4.91t2 + 5.0t -4 =0
    However I can't do this because using quadratic I get a negative. Therefore it's not an option.

    ATTEMPT 2: CAN I ASSUME v2=0?
    d=v1+v2/2 x t
    4=(5)+ (0)/2 x t
    t = 1.6s
    * But I guess I can't assume the V2=0

    Attempt 3: CAN I ASSUME v2=0?
    v2=v1+at
    0=5 +(-9.81)t
    t = 0.51 s.

    I know the answer is 0.53s, however how do I know whether I should use attempt 2 or attempt 3?
    I know to use attempt 3 because it's the right answer, but in my mind 1.6 s makes sense too.

    Please help!
     
  2. jcsd
  3. Jan 27, 2017 #2

    PeroK

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    If it hits the branch at ##5m/s## is that the initial velocity or the final velocity?

    Think about the motion that you are studying. Where does it start and where does it end?
     
  4. Jan 27, 2017 #3
    I believe it starts out of the sling at 5.0 m/s so thats v1. However it ends when it hits the branch? But it doesn't really, it will just bounce off and go in another direction.
     
  5. Jan 27, 2017 #4

    PeroK

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    The "final" velocity is the velocity at the end of the motion you are interested in. In this case, when it hits the branch. You are not asked to study the motion of the stone after that.

    The question says that the stone hits the branch at ##5m/s##. It does not say it leaves the slingshot at ##5m/s##.
     
  6. Jan 27, 2017 #5
    Thank you! To clarify: the motion I am interested in is the rock flying from the slingshot. Therefore can I assume that v1=0? Or no because v1 is the rock out of the slingshot and therefore it would be a positive value?
     
  7. Jan 27, 2017 #6

    PeroK

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    If ##v_1 = 0##, how far do you think the stone would go? You are not given ##v_1##. It's an unknown quantity.
     
  8. Jan 27, 2017 #7
    I understand that's in an unknown, I just assumed I could assume. Especially since my only two available equations have v1 in them.
    d=v1+v2/2 x t
    4=(unknown) + (5)/2 x t
    t = 1.6s (I know this is incorrect)

    v2=v1+at
    5=unknown +(-9.81)t
    t = -0.51 s. (I know this is correct but wrong sign)
     
  9. Jan 27, 2017 #8

    PeroK

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    Both those equations have two unknowns: ##v_1## and ##t##. So, you may need to think of another equation!
     
  10. Jan 27, 2017 #9
    So for each movement I should ask myself:
    What is the segment of movement? In this case it's the rock out of a slingshot.
    Therefore v1 is unknown because it's already launched out of the slingshot.
    V2 = given
    Because I am having difficulties knowing if v1=0 or v2=0 or neither
     
  11. Jan 27, 2017 #10

    PeroK

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    Yes. In this case ##v_1## is unknown and ##v_2 = 5m/s##.

    But, in this case, you are missing an important kinematic equation.
     
  12. Jan 27, 2017 #11
    I understand! I should solve for v1, then use a secondary equation to find t.
     
  13. Jan 27, 2017 #12

    PeroK

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    You can do it with the equations you have. But, there is another equation that would make things easier.
     
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