A particle is released from rest from a tower | vedclass

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(D) Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$,we get $S = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2S}{

Vedclass · Accepted Answer

$1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$

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(D) Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$,we get $S = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2S}{g}}$.For the first height $h$,the time taken is $t_1 = \sqrt{\frac{2h}{g}}$.For the first two heights (total distance $2h$),the time taken is $T_2 = \sqrt{\frac{2(2h)}{g}} = \sqrt{\frac{4h}{g}}$. Thus,the time taken for the second interval $h$ is $t_2 = T_2 - t_1 = \sqrt{\frac{4h}{g}} - \sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}}(\sqrt{2} - 1)$.For the first three heights (total distance $3h$),the time taken is $T_3 = \sqrt{\frac{2(3h)}{g}} = \sqrt{\frac{6h}{g}}$. Thus,the time taken for the third interval $h$ is $t_3 = T_3 - T_2 = \sqrt{\frac{6h}{g}} - \sqrt{\frac{4h}{g}} = \sqrt{\frac{2h}{g}}(\sqrt{3} - \sqrt{2})$.Taking the ratio $t_1 : t_2 : t_3$:$t_1 : t_2 : t_3 = \sqrt{\frac{2h}{g}} : \sqrt{\frac{2h}{g}}(\sqrt{2} - 1) : \sqrt{\frac{2h}{g}}(\sqrt{3} - \sqrt{2})$$t_1 : t_2 : t_3 = 1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$.

$A$ particle is released from rest from a tower of height $3h$. The ratio of times taken to fall through successive equal heights $h$,i.e.,$t_1 : t_2 : t_3$ is

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