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(D) Let the point $P$ be at a height $h$ from the ground. The equation of motion is $h = vt - \frac{1}{2}gt^2$.Rearranging this,we get $\frac{1}{2}gt^

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$2(\frac{v}{g}-x)$

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(D) Let the point $P$ be at a height $h$ from the ground. The equation of motion is $h = vt - \frac{1}{2}gt^2$.Rearranging this,we get $\frac{1}{2}gt^2 - vt + h = 0$.This is a quadratic equation in $t$,which has two roots $t_1$ and $t_2$,representing the times when the ball is at height $h$.Given $t_1 = x$,the sum of the roots is $t_1 + t_2 = \frac{-(-v)}{\frac{1}{2}g} = \frac{2v}{g}$.Therefore,$t_2 = \frac{2v}{g} - x$.The time taken to reach the point $P$ again from the first time it passed through $P$ is $t_2 - t_1 = (\frac{2v}{g} - x) - x = \frac{2v}{g} - 2x = 2(\frac{v}{g} - x)$.

$A$ ball projected vertically upwards with a velocity '$v$' passes through a point $P$ in its upward journey in a time of '$x$' seconds. From there,the time in which the ball again passes through the same point $P$ is

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