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 will be(take g as acceleration due to gravity).

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

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

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

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

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Solution

The correct option is C $\frac{g (t_{1} + t_{2})}{2}$
Let v be initial velocity of vertical projection and t be the time taken by the body to reach a height h from ground.
Here, $u = u, a = - g, s = h, t = t$
Using, $s = u t + \frac{1}{2} a t^{2}$ , we have
$h = u t + \frac{1}{2} (- g) t^{2}$ or $g t^{2} - 2 u t + 2 h = 0$
$∴ t = \frac{2 u \pm \sqrt{4 u^{2} - 4 g \times 2 h}}{2 g} = \frac{u \pm \sqrt{u^{2} - 2 g h}}{g}$
It means t has two values, i.e.,
$t_{1} = \frac{u + \sqrt{u^{2} - 2 g h}}{g}$
$t_{2} = \frac{u - \sqrt{u^{2} - 2 g h}}{g}$
$t_{1} + t_{2} = \frac{2 u}{g}$ or $u = \frac{g (t_{1} + t_{2})}{2}$ .

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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 will be(take g as acceleration due to gravity).

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