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Q.
A body is dropped from a height $H$. The time taken to cover second half of the journey is
Motion in a Straight Line
Solution:
The total time of journey
$-s=u t-\frac{1}{2} g t^{2}$
$\Rightarrow H=\frac{1}{2} g T^{2}$ ...(i)
$\frac{-H}{2}=u t-\frac{1}{2} g t^{2}$
$\Rightarrow T=\sqrt{\frac{2 H}{g}}$
$\Rightarrow \frac{H}{2}=\frac{1}{2} g t^{2}$
$\Rightarrow \frac{1}{2} g T^{2}=g t^{2} (\because u t=0)$
$\Rightarrow t-\frac{T}{\sqrt{2}}$
$\Rightarrow $ Second half time $=T-t=T-\frac{T}{\sqrt{2}}$
$=T\left(1-\frac{1}{\sqrt{2}}\right)=\sqrt{\frac{2 H}{g}}\left(1-\frac{1}{\sqrt{2}}\right)$
$=\sqrt{\frac{H}{g}}(\sqrt{2}-1)$
