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A body thrown vertically up to reach its maximum height in t seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) is

\[\begin {array} {1 1} (1)\;\sqrt 2 t & \quad (2)\;\bigg(1+ \frac{1}{\sqrt 2}\bigg)t \\ (3)\;\frac{3t}{2} & \quad (4)\;\frac{t}{\sqrt 2} \end {array}\]
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$(2)\;\bigg(1+ \frac{1}{\sqrt 2}\bigg)t$
answered Nov 7, 2013 by pady_1

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