A body is projected vertically upwards. The times corresponding to height $h$ while ascending and while descending are $t_{1}$ and $t_{2}$ respectively. Then the velocity of projection is ( $g$ is acceleration due to gravity)

g \sqrt{t_{1} t_{2}}

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

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\frac{g \sqrt{t_{1} t_{2}}}{2}

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

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Solution

The correct option is A $g \sqrt{t_{1} t_{2}}$
$h = u t - \frac{g t^{2}}{2}$
on solving the above quadratic we get
$t_{1, 2} = \frac{u \pm \sqrt{(u^{2} - 2 g h)}}{g}$
$t_{1} \times t_{2} = 2 h / g$
Also $V^{2} = 2 g h$
$v = g \sqrt{t_{1} t_{2}}$

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A body is projected vertically upwards. The times corresponding to height h while ascending and while descending are t1 and t2 respectively. Then the velocity of projection is (g is acceleration due to gravity)

The correct option is A g√t1t2h=ut−gt22on solving the above quadratic we gett1,2=u±√(u2−2gh)gt1×t2=2h/gAlso V2=2ghv=g√t1t2

A body is projected vertically upwards. The times corresponding to height $h$ while ascending and while descending are $t_{1}$ and $t_{2}$ respectively. Then the velocity of projection is ( $g$ is acceleration due to gravity)

The correct option is A $g \sqrt{t_{1} t_{2}}$
$h = u t - \frac{g t^{2}}{2}$
on solving the above quadratic we get
$t_{1, 2} = \frac{u \pm \sqrt{(u^{2} - 2 g h)}}{g}$
$t_{1} \times t_{2} = 2 h / g$
Also $V^{2} = 2 g h$
$v = g \sqrt{t_{1} t_{2}}$