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A body is projected vertically upwards at time $t=0$ and it is seen at a height $'H'$ at time $t_1$ and $t_2$ seconds during its flight. The maximum height attained is : (g is acceleration due to gravity)

\[\begin {array} {1 1} (1)\;\frac{g(t_2-t_1)^2}{8} & \quad (2)\;\frac{g(t_2-t_1)^2}{4} \\ (3)\;\frac{g(t_1+t_2)^2}{8} & \quad (4)\;\frac{g(t_2-t_1)^2}{4} \end {array}\]
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answered Nov 7, 2013 by pady_1

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