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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

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$g t_{1}$

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

$g t_{2}$

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

answer is D.

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Detailed Solution

If $t_{1}$ and $t_{2}$ are time of ascent and descent respectively then time of flight $T = t_{1} + t_{2} = \frac{2 u}{g}$ Þ $u = \frac{g (t_{1} + t_{2})}{2}$

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