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(C) Let the time taken by the body to fall from the maximum height point $C$ to point $B$ be $t^{\prime}$.Due to symmetry of motion, the time taken to

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$\frac{g(t_1+t_2)^2}{8}$

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(C) Let the time taken by the body to fall from the maximum height point $C$ to point $B$ be $t^{\prime}$.Due to symmetry of motion, the time taken to go from $B$ to $C$ is equal to the time taken to fall from $C$ to $B$, which is $t^{\prime}$.Given that the body is at height $H$ at times $t_1$ and $t_2$, we have $t_2 = t_1 + 2t^{\prime}$.Thus, $t^{\prime} = \frac{t_2-t_1}{2}$.The total time $T$ taken to reach the maximum height $C$ is $T = t_1 + t^{\prime} = t_1 + \frac{t_2-t_1}{2} = \frac{t_1+t_2}{2}$.The maximum height $H_{\max}$ attained is given by the formula $H_{\max} = \frac{1}{2}gT^2$.Substituting the value of $T$, we get $H_{\max} = \frac{1}{2}g\left(\frac{t_1+t_2}{2}\right)^2 = \frac{1}{2}g \cdot \frac{(t_1+t_2)^2}{4} = \frac{g(t_1+t_2)^2}{8}$.

$A$ body is projected vertically upwards at time $t=0$ and it is seen at a height $H$ at times $t_1$ and $t_2$ seconds during its flight. The maximum height attained is ($g$ is acceleration due to gravity).

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