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a small ball is released from rest on a smooth plane inclined at 30 degree to the horizontal.find the distance the particle has travelled down the plane when its speed is 6.3m/s. i've an idea to solve using this formula: v^2=u^2+2as but i refuse where we should use the angle given? please help me..
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You have to resolve the acceleration into a horizontal and vertical components so a = 9.81 sin(30) = 4.905 m/s² then , you'll use the equation you mentioned
tq ^_^
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