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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)
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Then, the velocity of projection will be (take $g$ as acceleration due to gravity)
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Verified Answer
The correct answer is:
$\frac{g\left(t_1+t_2\right)}{2}$
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
$\begin{aligned}& h=u t+\frac{1}{2}(-g) t^2 \text { or } g t^2-2 u t+2 h=0 \\& \therefore t=\frac{2 u \pm \sqrt{4 u^2-4 g \times 2 h}}{2 g}=\frac{u \pm \sqrt{u^2-2 g h}}{g}\end{aligned}$
It means $t$ has two values, i.e.
$\begin{aligned}& t_1=\frac{u+\sqrt{u^2-2 g h}}{g} \\
& t_2=\frac{u-\sqrt{u^2-2 g h}}{g} \\& t_1+t_2=\frac{2 u}{g} \text { or } u=\frac{g\left(t_1+t_2\right)}{2}\end{aligned}$
Here, $u=u, a=-g, s=h, t=t$
Using, $s=u t+\frac{1}{2} a t^2$, we have
$\begin{aligned}& h=u t+\frac{1}{2}(-g) t^2 \text { or } g t^2-2 u t+2 h=0 \\& \therefore t=\frac{2 u \pm \sqrt{4 u^2-4 g \times 2 h}}{2 g}=\frac{u \pm \sqrt{u^2-2 g h}}{g}\end{aligned}$
It means $t$ has two values, i.e.
$\begin{aligned}& t_1=\frac{u+\sqrt{u^2-2 g h}}{g} \\
& t_2=\frac{u-\sqrt{u^2-2 g h}}{g} \\& t_1+t_2=\frac{2 u}{g} \text { or } u=\frac{g\left(t_1+t_2\right)}{2}\end{aligned}$
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