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A ball is thrown vertically upwards from the top of a tower. Velocity at a point 'h' m vertically below the point of projection is twice the download velocity at a point 'h' m vertically above the point of projection. The maximum height reached by the ball above the top of the tower is

$\begin {array} {1 1} (1)\;\large\frac{4}{3}h & \quad (2)\;\large\frac{5h}{3} \\ (3)\;3\: h & \quad (4)\;2\: h \end {array}$

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(2) $ \large\frac{5h}{3}$
answered Nov 7, 2013 by pady_1 1 flag
Please explain the answer briefly

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