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A body is projected vertically up. What ...

A body is projected vertically up. What is the distance covered in its last second of upward motion? `(g=10 m//s^(2))`

Text Solution

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The distance covered by the object in its last second of its upward motion is equal to the distance covered in the first second of its downward motion.
Hence `s=1//2 gt^(2)=1//2 xx 10 xx 1=5 m`
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