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A dancer leaps off the floor with her centre of mass having a velocity of 5 m/s making an angle of `theta =37^(@)` to the horizontal. At the top of the trajectory the dancer has her legs stretched so that the centre of mass gets closer to head by a vertical distance of `0.25 m`. By how much does the head rises vertically from its initial position ? `[ sin 37^(@) = (3)/(5)]` |
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Answer» Correct Answer - 0.2 m |
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