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A stone is thrown vertically upward.

  1. Jan 22, 2014 #1
    1. The problem statement, all variables and given/known data

    A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B, 3.00 m higher than A, with speed v/2. Calculate (a) the speed v and (b) the maximum height reached by the stone above the point B.

    2. Relevant equations

    Kinematic Equations

    v = v(knot) + a(t)

    V^2 = v(knot)^2 + 2a (x - x(knot))

    x - x(knot) = v(knot)t + 0.5a(t)^2

    3. The attempt at a solution

    Because I'm not given time in the problem statement, I believe I start with Kinematic eq two. If I want to find speed at point A, I don't know how to represent that as an x value. I know "a" is "-9.8" but am not given v(knot)^2. I'm left with too many variables.

    Thank you.
  2. jcsd
  3. Jan 22, 2014 #2
    You have chosen the correct formula.

    Use that formula from initial speed v to speed v/2.
  4. Jan 22, 2014 #3
    But what is my v(knot) and x(knot)? Thank you.
  5. Jan 22, 2014 #4
    You take point A as the zero reference for distance.
  6. Jan 22, 2014 #5
    BTW does not v[itex]_{o}[/itex] look better?
  7. Jan 22, 2014 #6
    V^2 = v(knot)^2 + 2a (A-x(knot))?

  8. Jan 22, 2014 #7


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    Homework Helper

    Welcome to PF!

    Yes, use the second kinematic equations. Substitute all values given, write XA and xB for the positions. You know that xB=3+xA.

  9. Jan 22, 2014 #8
    I am going to write your 'v(knot)' as v[itex]_{o}[/itex].

    I repeat. Take point A as your zero reference for the distance. Hence the distance at A would have the value of 0.
  10. Jan 22, 2014 #9
    Or follow the suggestion of ehild.
  11. Jan 22, 2014 #10
    Thank you.

    But does that not still leave me with V^2 = v(knot)^2 + 2(-9.8)(3-A)?

  12. Jan 22, 2014 #11


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    Homework Helper

    You have two equations.

    If x =A v =V, given.
    If x=B=3+A, v= V/2.

    Show these equations.

    Last edited: Jan 22, 2014
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