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Q.
A ball dropped from the top of tower falls first half height of tower in $10\, s$. The total time spend by ball in air is [Take $g=10\, m / s ^{2}$ ]
Motion in a Straight Line
Solution:
$\frac{-H}{2}=u t-\frac{1}{2} g \times 10^{2}$
$\Rightarrow H=g \times 10^{2}$
$\Rightarrow -H=-\frac{1}{2} g t^{2}$ (Full journey)
$g \times 10^{2}=\frac{1}{2} g t^{2}$
$\Rightarrow t^{2}=200$
$\Rightarrow t=10 \sqrt{2} s$
$\Rightarrow t=10 \times 1.414\,s$
$=14.14\, s =t$
