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A body projected vertically upwards crosses a point twice in its journey at a height h just after $t_1$ and $t_2$ seconds. Maximum height reached by the body is :

\[\begin {array} {1 1} (a)\;\frac{g}{4} (t_1+t_2)^2 & \quad (b)\;g\bigg(\frac{t_1+t_2}{4}\bigg)^2 \\ (c)\;2g \bigg(\frac{t_1+t_2}{4}\bigg)^2 & \quad (d)\;\frac{g}{4} (t_1t_2) \end {array}\]

1 Answer

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$ (C)\;2g \bigg(\frac{t_1+t_2}{4}\bigg)^2$
Hence C is the correct answer.
answered Mar 24, 2014 by meena.p

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