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A body is projected vertically upwards. The times corresponding to height $h$, while ascending and descending are $t_{1}$ and $t_{2}$, respectively. Then, the velocity of projection is ( $g$ is acceleration due to gravity)
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Verified Answer
The correct answer is:
$\frac{g\left(t_{1}+t_{2}\right)}{2}$
Let $u$ be the initial vertical velocity of the body by which it is projected vertically upward.
If $t$ be the time taken by the body to reach a height $h$ from the point of projection, then using equation of vertical motion,
$$
\begin{aligned}
h &=u t-\frac{1}{2} g t^{2} \\
\Rightarrow \quad & g t^{2}-2 u t+2 h=0 \\
\Rightarrow \quad t &=\frac{2 u \pm \sqrt{4 u^{2}-4 g \times 2 h}}{2 g} \\
&=\frac{u \pm \sqrt{u^{2}-2 g h}}{g}
\end{aligned}
$$
It means $t$ has two values which is already given in question,
$$
t_{1}=\frac{u+\sqrt{u^{2}-2 g h}}{g}
$$
$\begin{aligned} & \text { and } & t_{2} &=\frac{u-\sqrt{u^{2}-2 g h}}{g} \\ \therefore & t_{1}+t_{2} &=\frac{2 u}{g} \\ \Rightarrow & u &=\frac{g\left(t_{1}+t_{2}\right)}{2} \end{aligned}$
If $t$ be the time taken by the body to reach a height $h$ from the point of projection, then using equation of vertical motion,
$$
\begin{aligned}
h &=u t-\frac{1}{2} g t^{2} \\
\Rightarrow \quad & g t^{2}-2 u t+2 h=0 \\
\Rightarrow \quad t &=\frac{2 u \pm \sqrt{4 u^{2}-4 g \times 2 h}}{2 g} \\
&=\frac{u \pm \sqrt{u^{2}-2 g h}}{g}
\end{aligned}
$$
It means $t$ has two values which is already given in question,
$$
t_{1}=\frac{u+\sqrt{u^{2}-2 g h}}{g}
$$
$\begin{aligned} & \text { and } & t_{2} &=\frac{u-\sqrt{u^{2}-2 g h}}{g} \\ \therefore & t_{1}+t_{2} &=\frac{2 u}{g} \\ \Rightarrow & u &=\frac{g\left(t_{1}+t_{2}\right)}{2} \end{aligned}$
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