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(D) Let the initial velocity at point $A$ be $u$ and the velocity at point $B$ be $v$. The time taken to reach $B$ from $A$ is $t_1$. Using the equati

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$\frac{1}{2} gt_1t_2$

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(D) Let the initial velocity at point $A$ be $u$ and the velocity at point $B$ be $v$. The time taken to reach $B$ from $A$ is $t_1$. Using the equation $v = u - gt_1$,we have $v = u - gt_1$. After reaching the maximum height,the particle returns to $B$ and then to $A$. The time taken to return from $B$ to the ground (point $A$) is $t_2$. At point $B$,the velocity is $v$ (upwards). When it returns to $B$ on the way down,the velocity is $-v$ (downwards). The time taken to travel from $B$ to the ground is $t_2$. Using $v = u + at$,we have $-u = v - gt_2$,which implies $v = gt_2 - u$. Equating the two expressions for $v$: $u - gt_1 = gt_2 - u$,which gives $2u = g(t_1 + t_2)$,or $u = \frac{1}{2}g(t_1 + t_2)$. The height $h$ of point $B$ from the ground is given by $h = ut_1 - \frac{1}{2}gt_1^2$. Substituting $u$: $h = \left[\frac{1}{2}g(t_1 + t_2)\right]t_1 - \frac{1}{2}gt_1^2 = \frac{1}{2}gt_1^2 + \frac{1}{2}gt_1t_2 - \frac{1}{2}gt_1^2 = \frac{1}{2}gt_1t_2$.

$A$ particle is projected vertically upwards from a point $A$ on the ground. It takes $t_1$ time to reach a point $B$ but it still continues to move up. If it takes $t_2$ time to reach the ground from point $B$,then the height of point $B$ from the ground is

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