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(B) Let the height of the point be $h$. Using the equation of motion $h = ut - \frac{1}{2}gt^2$,we have $h = ut_1 - \frac{1}{2}gt_1^2$.Let $t_2$ be th

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$2\left(\frac{u}{g}-t_1\right)$

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(B) Let the height of the point be $h$. Using the equation of motion $h = ut - \frac{1}{2}gt^2$,we have $h = ut_1 - \frac{1}{2}gt_1^2$.Let $t_2$ be the total time taken to reach the same point during the return journey. Then $h = ut_2 - \frac{1}{2}gt_2^2$.Equating the two expressions for $h$: $ut_1 - \frac{1}{2}gt_1^2 = ut_2 - \frac{1}{2}gt_2^2$.Rearranging the terms: $u(t_1 - t_2) = \frac{1}{2}g(t_1^2 - t_2^2) = \frac{1}{2}g(t_1 - t_2)(t_1 + t_2)$.Since $t_1 
eq t_2$,we can divide by $(t_1 - t_2)$ to get $u = \frac{1}{2}g(t_1 + t_2)$,which implies $t_1 + t_2 = \frac{2u}{g}$.The time interval between the two passages is $\Delta t = t_2 - t_1$.From $t_2 = \frac{2u}{g} - t_1$,we get $\Delta t = \left(\frac{2u}{g} - t_1ight) - t_1 = \frac{2u}{g} - 2t_1 = 2\left(\frac{u}{g} - t_1ight)$.

$A$ body is projected vertically upwards with a velocity $u$. It passes through a certain point above the ground after $t_1$ seconds. The time interval after which the body passes through the same point during the return journey is

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