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(D) Let $u$ be the initial velocity of the particle at point $A$.The total time taken to reach the ground again is $T = t_1 + t_2$.Using the equation

Vedclass · Accepted Answer

$\frac{1}{2} g t_1 t_2$

Vedclass · Answer

(D) Let $u$ be the initial velocity of the particle at point $A$.The total time taken to reach the ground again is $T = t_1 + t_2$.Using the equation of motion for the total flight: $0 = u(t_1 + t_2) - \frac{1}{2} g(t_1 + t_2)^2$.Solving for $u$: $u = \frac{1}{2} g(t_1 + t_2)$.The height $h$ of point $B$ from the ground is given by the displacement at time $t_1$: $h = u t_1 - \frac{1}{2} g t_1^2$.Substituting the value of $u$: $h = \left[ \frac{1}{2} g(t_1 + t_2) \right] t_1 - \frac{1}{2} g t_1^2$.Simplifying the expression: $h = \frac{1}{2} g t_1^2 + \frac{1}{2} g t_1 t_2 - \frac{1}{2} g t_1^2$.Therefore,$h = \frac{1}{2} g t_1 t_2$.

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

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