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(C) Let the body reach the maximum height $C$ at time $T$. Due to symmetry,the time taken to go from $C$ to $B$ is the same as the time taken to go fr

Vedclass · Accepted Answer

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

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(C) Let the body reach the maximum height $C$ at time $T$. Due to symmetry,the time taken to go from $C$ to $B$ is the same as the time taken to go from $B$ to $C$. Let this time be $t^{\prime}$.Thus,$t_2 = T + t^{\prime}$ and $t_1 = T - t^{\prime}$.Adding these equations: $t_1 + t_2 = 2T$,so $T = \frac{t_1 + t_2}{2}$.Subtracting these equations: $t_2 - t_1 = 2t^{\prime}$,so $t^{\prime} = \frac{t_2 - t_1}{2}$.The maximum height $H_{\max}$ is the distance covered in time $T$ starting from rest at point $C$ (downward motion).$H_{\max} = \frac{1}{2} g T^2 = \frac{1}{2} g \left( \frac{t_1 + t_2}{2} \right)^2 = \frac{g(t_1 + t_2)^2}{8}$.

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

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