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Vertical Projectile Motion

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A mountain climber stands at the top of a 50m cliff that overhangs a calm pool. She throws two stones vertically download 1s apart and observe that the two stones reach the water simultaneously after a while. The first stone was thrown at an initial speed of 2m[tex]^{-1}[/tex].s

    Calculate the initial speed at which she threw the second stone. Ignore the effects of friction


    2. Relevant equations



    3. The attempt at a solution
    Vf[tex]^{2}[/tex]=V[tex]_{i}[/tex][tex]^{2}[/tex] + a[tex]\Delta[/tex]y
    =2(2)[tex]^{2}[/tex]=2(9.8)(50)=31.37
    Vf=5.6m.s[tex]^{-1}[/tex]

    V[tex]_{f}[/tex]=v[tex]_{i}[/tex]+at
    5.6=2+9.8t
    t=0.367....
    t=0.37s

    I am not sure if im right...because now i have to -1 second from the t above hence giving me a negative second to enter in a formula to find the initial velocity of the stone...

    Can someone help me...the model answer gives me an anser of 15.2m/s downwards...how can that be? surely it is wrong
     
  2. jcsd
  3. Sep 8, 2009 #2
    Find how long it takes the first rock to reach the water, that also will let you easily find the time the second rock was in the air. Then just solve for Vo.
     
  4. Sep 9, 2009 #3
    i did an i get 0.37s...and if the the second stone was released 1 s later then how can that be?
     
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