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Assistance with Projectiles

  1. Oct 1, 2005 #1
    A stone is projected from a point O on a cliff with a speed of of 20ms^-1 at an angle of elevation of 30. T seconds later the angle of depression of the stone from O is 45. Find the value of T.

    I assumed that when the stone is at depression of 45 degree to O, it has reached the horizontal plane again, hence displacement is 0. so I used s = ut + 0.5at^2.
    s=0
    a=-9.8
    u=20sin30
    therefore t= 2.04, which is apparabtly wrong.

    Please help.
     
  2. jcsd
  3. Oct 1, 2005 #2

    Päällikkö

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    You've assumption of the angle is wrong.
    [tex]\tan \theta = \frac{v_y}{v_x}[/tex]
     
  4. Oct 1, 2005 #3
    Could you expand on that please?
     
  5. Oct 1, 2005 #4

    Päällikkö

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  6. Oct 1, 2005 #5
    But how can I show that angle of depression is 45 degrees on that diagram?
     
  7. Oct 1, 2005 #6

    Päällikkö

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    You'll want theta = -45o (depending on the positive/negative directions you've chosen).

    On the other hand, as tan -45o = -1, vx = -vy (at the desired instant).
     
    Last edited: Oct 1, 2005
  8. Oct 1, 2005 #7
    Sorry Im confused! Could you go through the question please?
     
  9. Oct 1, 2005 #8

    Päällikkö

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    Positive/negative directions:
    +y
    |
    |
    ----+x

    [tex]\tan \theta = \frac{v_y}{v_x}[/tex]
    We also know that
    [tex]v = v_0+at[/tex]


    [tex]\tan \theta = \frac{v_{0y}-gt}{v_{0x}}[/tex]
     
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