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Question

(A) Let the height of the tower be $h$ and the initial speed be $u$. Taking the downward direction as positive,the equation of motion is $h = ut + \fr

Vedclass · Accepted Answer

$\sqrt{t_1 t_2}$

Vedclass · Answer

(A) Let the height of the tower be $h$ and the initial speed be $u$. Taking the downward direction as positive,the equation of motion is $h = ut + \frac{1}{2}gt^2$.For upward projection,the initial velocity is $-u$: $h = -ut_1 + \frac{1}{2}gt_1^2 \Rightarrow \frac{1}{2}gt_1^2 - ut_1 - h = 0$. Solving for $t_1$ (taking positive root): $t_1 = \frac{u + \sqrt{u^2 + 2gh}}{g}$.For downward projection,the initial velocity is $+u$: $h = ut_2 + \frac{1}{2}gt_2^2 \Rightarrow \frac{1}{2}gt_2^2 + ut_2 - h = 0$. Solving for $t_2$ (taking positive root): $t_2 = \frac{-u + \sqrt{u^2 + 2gh}}{g}$.If the body is dropped,$u = 0$,so $h = \frac{1}{2}gt^2 \Rightarrow t = \sqrt{\frac{2h}{g}}$.Multiplying $t_1$ and $t_2$: $t_1 t_2 = \left(\frac{\sqrt{u^2 + 2gh} + u}{g}\right) \left(\frac{\sqrt{u^2 + 2gh} - u}{g}\right) = \frac{(u^2 + 2gh) - u^2}{g^2} = \frac{2gh}{g^2} = \frac{2h}{g} = t^2$.Therefore,$t = \sqrt{t_1 t_2}$.

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