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A stone is thrown verticall upward

  1. Feb 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Question Details:
    9. Problem: A stone is thrown verticall upward with a speed of 12.0 m/s from the edge of a cliff 70.0m high. How much later does it reach the bottom of the cliff? What is its speed just before hitting the ground? What total distance did it travel?

    time=
    speed=
    distance=



    2. Relevant equations
    y = yo + vot + 1/2 at2
    Δy = Vot + 1/2 gt2
    v2 = vo2 + 2g Δy
    v = vo + gt


    3. The attempt at a solution

    v = vo + gt
    0=12 + (-9.8)t
    -12= -9.8t
    1.22=t

    V2 = Vo2 +2g Δy
    02 = (12)2 +2 (-9.8) Δy
    02 = 144 + (-19.6) Δy
    -144 = -19.6 Δy
    7.35=Δy

    Δy= Vot + 1/2 gt2
    77.35 = o +1/2 (9.8)t2
    77.35 = o +4.9 t2
    15.79 = t2
    3.97 = t

    1.22 t + 3.97 t = 5.19 t

    distance is 7.35m x 2 + 70m = 84.7 m

    V= Vo = at
    = -12 + 9.8 (5.2)
    = -12 + 50.96
    = 38.9 m/s

    am I close or totally lost? Help please!
     
  2. jcsd
  3. Feb 14, 2009 #2
    "Vertically upward", as stated in your scenario, fails to define any deviation from vertical therefore, there would be no reason for the stone to move in the horizontal direction. This would prevent the stone from ever falling over the 70 meter cliff. Your scenario is missing an angle of trajectory.
     
  4. Feb 14, 2009 #3
    That is the whole problem. Where do I go next then or what is the right equation?
     
  5. Feb 14, 2009 #4

    tiny-tim

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    Hi ironhawk316! :smile:

    Why are you doing this in stages?

    Use s = v0t + at2/2 starting at the start and finishing at the finish. :wink:

    (oh … and please don't use yellow!)
     
  6. Feb 14, 2009 #5
    Without an angle of trajectory provided, there won't be any free-falling of the stone over the 70 meter cliff. Since they want the values as per the 70 meter free-fall over the cliff, they must provide the angle of trajectory for the stone.

    "Vertical upward" or "vertically upward" fails to provide any X (horizontal) direction and they fail to provide any deviation from vertical. The stone will simply fall right back to where it was thrown unless they provide you with an angle of trajectory. The scenario stated is lacking crucial info.
     
  7. Feb 14, 2009 #6
    Maybe the stone-thrower's arm is extended over the edge of the cliff. I think that might be what the problem wants you to assume.
     
  8. Oct 1, 2010 #7
    why don't you use v^2 - u^2 = 2as directly after getting displacement?
    s = displacement
    u = initial velocity
    v = final velocity
    a = gravity = -9.8

    getting displacement = 12 x 1.2
    = 14.4m
    single vertical distance of rock = 7.2m

    s = 77.2
    u = 0
    v = ?
    a = -9.8
    t = ?

    then directly use v^2 - u^2 = 2as
    v^2 - 0 = 2(-9.8)(77.2)
    v^2 - 0 = 2(-9.8)(77.2)
    v^2 = 1513.12
    v = 38.9m/s

    this should be easier than what you did above~
     
  9. Jul 30, 2011 #8

    :approve:
    i need help!!!
    a stone is thrown upward in a vertical manner..
    wat is its acceleration while it is going up!!
     
  10. Jul 30, 2011 #9

    tiny-tim

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    Last edited by a moderator: Apr 26, 2017
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