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Question

(B) Let the initial velocity be $u$. The total time of flight is $T = t_1 + t_2$. Since the ball returns to the ground,the total displacement is $0$.

Vedclass · Accepted Answer

$\frac{1}{2} gt_1 t_2$

Vedclass · Answer

(B) Let the initial velocity be $u$. The total time of flight is $T = t_1 + t_2$. Since the ball returns to the ground,the total displacement is $0$. Using the equation of motion $S = ut + \frac{1}{2}at^2$ with $a = -g$:$0 = u(t_1 + t_2) - \frac{1}{2}g(t_1 + t_2)^2$$u = \frac{1}{2}g(t_1 + t_2)$Now,the height $h$ reached at time $t_1$ is given by:$h = ut_1 - \frac{1}{2}gt_1^2$Substituting the value of $u$:$h = \left[\frac{1}{2}g(t_1 + t_2)\right]t_1 - \frac{1}{2}gt_1^2$$h = \frac{1}{2}gt_1^2 + \frac{1}{2}gt_1t_2 - \frac{1}{2}gt_1^2$$h = \frac{1}{2}gt_1t_2$

$A$ ball is projected vertically upwards from the ground. It reaches a height $h$ in time $t_1$,continues its motion,and then takes a time $t_2$ to reach the ground. The height $h$ in terms of $g, t_1$,and $t_2$ is ($g =$ acceleration due to gravity).

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