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mathslover

A particle is projected with speed \(\sqrt{3} i\cap + j\cap +k\cap\) from the origin. If the highest point of trajectory , the velocity vector makes and angle \(\theta\) with n-axis, then find \(\tan \theta\)

  • one year ago
  • one year ago

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  1. mathslover
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    I don't know how to represent i cap

    • one year ago
  2. mathslover
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    @sauravshakya

    • one year ago
  3. mathslover
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    |dw:1349108807115:dw|

    • one year ago
  4. Yahoo!
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    Three Co-orDinates...hm......if it is Two Cordinate tanx = vy/vx @highest point of trajectory vertical component of velocity is 0

    • one year ago
  5. mathslover
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    \[u\sin \theta = 0\]

    • one year ago
  6. mathslover
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    Now? sin theta = 0?

    • one year ago
  7. Yahoo!
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    if this is Two-Cordinate then usinx =0........Since it is a Three Coordinate...) i dont thhink...so...and i have nt studied....Three Coordinate Plane...

    • one year ago
  8. mathslover
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    hmn no problem. @experimentX @sauravshakya @helder_edwin any ideas?

    • one year ago
  9. mathslover
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    @Jemurray3 @mukushla @ghazi @ganeshie8

    • one year ago
  10. ghazi
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    @mathslover what does n axis refer to? this is supposed to be a three dimensional coordinate system

    • one year ago
  11. ghazi
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    and theta= 30 degrees

    • one year ago
  12. mathslover
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    Well the question dont tells this .... I am also confused , also in starting it is written : Treat vertically upward as K|dw:1349109611966:dw|

    • one year ago
  13. ghazi
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    i mean \[\tan \theta = \frac{ 1 }{ \sqrt3 }\]

    • one year ago
  14. mathslover
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    You're correct @ghazi but HOW?

    • one year ago
  15. ghazi
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    see if you consider a three dimensional system check out the spherical coordinate system on google you'll find theta is the angle between x and y axis now \[\tan \theta= \frac{ y }{ x }=\frac{ 1 }{ \sqrt3 }\] x and y are the intercepts

    • one year ago
  16. mathslover
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    Oh very easy it was, I feel it now. Thanks a ton @ghazi

    • one year ago
  17. ghazi
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    :)

    • one year ago
  18. amistre64
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    this is what i was thinking |dw:1349110013381:dw|

    • one year ago
  19. ghazi
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    @amistre64 nice work, @mathslover this is what i was talking about .....

    • one year ago
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