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(B) Let the initial velocity be $u$. At maximum height $h$,the final velocity is $0$.Using $v = u - gt$,we get $0 = u - gt$,so $u = gt$.The maximum he

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$\left(1+\frac{1}{\sqrt{2}}\right) t$

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(B) Let the initial velocity be $u$. At maximum height $h$,the final velocity is $0$.Using $v = u - gt$,we get $0 = u - gt$,so $u = gt$.The maximum height is $h = \frac{u^2}{2g} = \frac{(gt)^2}{2g} = \frac{gt^2}{2}$.The body reaches the maximum height in time $t$.While returning,the body starts from rest at height $h$ and falls to height $h/2$. The distance covered is $s = h - h/2 = h/2$.Using $s = \frac{1}{2}gt'^2$,where $t'$ is the time taken to fall from maximum height to $h/2$:$\frac{h}{2} = \frac{1}{2}gt'^2 \implies t'^2 = \frac{h}{g} = \frac{gt^2/2}{g} = \frac{t^2}{2}$.Thus,$t' = \frac{t}{\sqrt{2}}$.The total time from projection is $T = t + t' = t + \frac{t}{\sqrt{2}} = \left(1 + \frac{1}{\sqrt{2}}\right) t$.

$A$ body is thrown vertically upwards and reaches its maximum height in $t$ seconds. The total time from the time of projection to reach a point at half of its maximum height while returning (in seconds) is:

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