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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 maximu

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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 falls from height $h$ to $h/2$. The distance covered is $h/2$.Using $s = ut + \frac{1}{2}at^2$ for the downward motion starting from rest at the maximum height:$\frac{h}{2} = 0 + \frac{1}{2}g(t')^2$, where $t'$ is the time taken to fall from the maximum height to half the maximum height.Substituting $h = \frac{gt^2}{2}$, we get $\frac{gt^2}{4} = \frac{1}{2}g(t')^2$.Solving for $t'$, we get $(t')^2 = \frac{t^2}{2}$, so $t' = \frac{t}{\sqrt{2}}$.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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