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A strange kind of straight line kinematics with uniform acceleration.

  1. Oct 17, 2012 #1
    1. The problem statement, all variables and given/known data
    A ball is thrown upward from a building 30 meters tall and misses the edge on its way down, hitting the ground 6 seconds after it was thrown.

    A) With what speed was it thrown?
    B) What is its highest height?


    2. Relevant equations
    Vf = Vi + at
    x = 1/2(Vi + Vf)
    x = Vi t + 1/2(at^2)
    Vf^2 = Vi^2 + 2ax

    Vf = final velocity
    Vi = initial velocity
    a = acceleration
    x = distance
    t = time

    3. The attempt at a solution
    So, it is a two part problem, the upward motion, and then the downward.
    I will label the distance from being thrown to the highest point of the arc as x1, which is also the same distance down to being level with where it was thrown from, and then x2 is 30 meters.

    Thus goes that to find time, I need to use x = Vi t + 1/2(at^2).
    xtotal = x1+30
    Vi = 0
    t2 = 6 - t1
    a = 10 (lets make positive downward)

    So here I am stumped by the fact that I do not have three known variable to make use of the equation. How am I supposed to proceed?
     
  2. jcsd
  3. Oct 17, 2012 #2

    lewando

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    Gold Member

    Make use of this form of the equation.

    x = x0 + v0t + 0.5at2

    [edit- you are trying to find v0]
     
    Last edited: Oct 17, 2012
  4. Oct 17, 2012 #3
    Yeah... so we have two missing variable then. An X and Vo :O
     
  5. Oct 17, 2012 #4

    lewando

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    Gold Member

    If the x-axis represents height, you know x0 and you know the final x. v0 is the only unknown. Totally solvable.
     
  6. Oct 17, 2012 #5
    If I take the base of the skyscraper as the start of the x axis, the ball is thrown up from 30m.

    x0 = 30m
    xfinal= 0

    That completely ignores the up and then down arc of the ball, what am I not understanding?
     
  7. Oct 17, 2012 #6

    lewando

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    The up and down arc is not relevant. Let the equation do the work.

    [edit-- "a = 10 (lets make positive downward)". If +x is pointing up, then a should be negative.]
     
    Last edited: Oct 17, 2012
  8. Oct 17, 2012 #7
    So...

    x = x0 + v0t + 0.5at2
    -30 = V0 + 0.5*(-10*62)
    -30 = V0 - 180
    150 = V0

    Is this it?
     
  9. Oct 17, 2012 #8

    lewando

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    Close! You dropped the "t" from v0.
     
  10. Oct 17, 2012 #9
    Oh, oops.

    x = x0 + v0t + 0.5at2
    -30 = V0*6 + 0.5*(-10*62)
    -30 = V0*6 - 180
    150 = V0*6
    V0 = 25

    Alright, great, thanks. The second part I can easily do on my own. :)
     
  11. Oct 17, 2012 #10

    lewando

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    Gold Member

    Good!
     
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