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a stone is thrown vertically upwards with a velocity of 40m/s and is caught back calculate the maxiimum height reached by the stone what is the net displacement and total distance covered by the stone |
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Answer» = 0U = 40 m/s a = -g m/s² = -10 m/s² 0 = (40)² -2 × 10 × S S = 80 m .hence MAXIMUM hight is reached by stone = 80 m .now, stone comes back to INTIAL POINT. hence total distance covered by stone = 80 ( in upward motion ) + 80( in DOWNWARD motion ) = 160 m . |
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