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El_Psy_Congroo!!

  • 3 years ago

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  1. El_Psy_Congroo!!
    • 3 years ago
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    |dw:1350258757152:dw|

  2. El_Psy_Congroo!!
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    |dw:1350258824409:dw|

  3. El_Psy_Congroo!!
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  4. El_Psy_Congroo!!
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  5. El_Psy_Congroo!!
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    the time taken by the particle to reach bottom is given by \[ s = ut +{1 \over 2}at^2, u=0, \\ s = {1 \over 2}at^2\\ a = g \cos 60 = {g \over 2}\\ 2h = {g \over 4}t^2 \\ t = \sqrt{8h \over g}\]

  6. El_Psy_Congroo!!
    • 3 years ago
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    |dw:1350259185816:dw| we use this formula \[ s = ut + {1 \over 2}at^2 = {1 \over 2} {\sqrt 3 \over 4}g {8h \over g} = \sqrt 3h\]

  7. El_Psy_Congroo!!
    • 3 years ago
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    a) \[ d= \sqrt 3 h\] b) \[ a=\sqrt 3 g\over 4\] c) \[ t = \sqrt{8h \over g} \]

  8. El_Psy_Congroo!!
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    |dw:1350265733178:dw|

  9. El_Psy_Congroo!!
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  10. RolyPoly
    • 3 years ago
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    Why is s = 2h?

  11. El_Psy_Congroo!!
    • 3 years ago
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    |dw:1350266031005:dw|

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