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Q.
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).
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, $s = ut +\frac{1}{2} a t^2$, we have
$
\begin{array}{l}
h = ut +\frac{1}{2}(- g ) t ^2 \text { or } gt ^2-2 ut +2 h =0 \\
\therefore 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 }
\end{array}
$
It means t has two values, i.e.,
$
\begin{array}{l}
t _1=\frac{ u +\sqrt{ u ^2-2 gh }}{ g } \\
t _2=\frac{ u -\sqrt{ u ^2-2 gh }}{ g } \\
t _1+ t _2=\frac{2 u }{ g } \text { or } u =\frac{ g \left( t _1+ t _2\right)}{2}
\end{array}
$