A particle is projected upwards. The times corresponding to height $h$ while ascending and while descending are $t_{1}$ and $t_{2}$ respectively. The velocity of projection will be

g t_{1}

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g t_{2}

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g t (t_{1} + t_{2})

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\frac{g (t_{1} + t_{2})}{2}

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Solution

The correct option is C $\frac{g (t_{1} + t_{2})}{2}$
Let $u$ be initial velocity, as for both time $t_{1} & t_{2}$ displacement is same $h$
$h = u t - \frac{1}{2} g t^{2} o r t = \frac{u \pm \sqrt{u^{2} - 4 (g / 2) h}}{g}$
$t_{1} = \frac{u - \sqrt{u^{2} - 4 (g / 2) h}}{g} a n d t_{2} = \frac{u + \sqrt{u^{2} - 4 (g / 2) h}}{g}$
$t_{1} + t_{2} = \frac{2 u}{g}$ $o r u = \frac{g (t_{1} + t_{2})}{2}$

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