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dw:1350258757152:dw
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the time taken by the particle to reach bottom is given by \[ s = ut +{1 \over 2}at^2, u=0, \\ s = {1 \over 2}at^2\\ a = g \cos 60 = {g \over 2}\\ 2h = {g \over 4}t^2 \\ t = \sqrt{8h \over g}\]
 one year ago

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dw:1350259185816:dw we use this formula \[ s = ut + {1 \over 2}at^2 = {1 \over 2} {\sqrt 3 \over 4}g {8h \over g} = \sqrt 3h\]
 one year ago

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a) \[ d= \sqrt 3 h\] b) \[ a=\sqrt 3 g\over 4\] c) \[ t = \sqrt{8h \over g} \]
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