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(C) At maximum height $h$,the final velocity is $0$. Using $v = u - gt$,we get $0 = u - gt$,so $u = gt$. The time to reach maximum height is $t = \fra

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$\frac{3h}{4}$

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(C) At maximum height $h$,the final velocity is $0$. Using $v = u - gt$,we get $0 = u - gt$,so $u = gt$. The time to reach maximum height is $t = \frac{u}{g}$.The maximum height $h$ is given by $h = \frac{u^2}{2g} = \frac{(gt)^2}{2g} = \frac{1}{2}gt^2$.Now,we calculate the height $h'$ covered in time $t' = \frac{t}{2}$ using the equation of motion $s = ut - \frac{1}{2}gt^2$:$h' = u(\frac{t}{2}) - \frac{1}{2}g(\frac{t}{2})^2$Substitute $u = gt$:$h' = (gt)(\frac{t}{2}) - \frac{1}{2}g(\frac{t^2}{4})$$h' = \frac{gt^2}{2} - \frac{gt^2}{8}$$h' = \frac{4gt^2 - gt^2}{8} = \frac{3gt^2}{8}$Since $h = \frac{1}{2}gt^2$,then $gt^2 = 2h$.Substituting this into the expression for $h'$:$h' = \frac{3(2h)}{8} = \frac{3h}{4}$.

$A$ body is projected vertically upwards with a velocity of $10\,m/s.$ It reaches the maximum vertical height $h$ in time $t.$ In $\frac{t}{2}$,the height covered is

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