Question

A particle at rest is dropped vertically from a height the time taken by it to fall successive distance of 1 metre it will then be in the ratio of difference in the square root of integers how to get this full answer I want full solution

Answer 1

Explanation:

hope this helps you to understand

Answer 2

Answer :

✴ Correct Question :

A particle is released from rest from a tower of height h. The ratio of the intervals of time to cover n successive distance of 1m is...

✴ Solution :

✒ Let t1, t2, t3, ... tn be the interval for n successive equal heights (1m) covered during the free fall of the particle.

Then...

$:\implies\sf\:1=\dfrac{1}{2}g{t_1}^2\\ \\ :\implies\bf\:t_1=\sqrt{\dfrac{2}{g}}\rightarrow\:(1)\\ \\ :\implies\sf\:2=\dfrac{1}{2}g{(t_1+t_2)}^2\\ \\ :\implies\bf\:t_1+t_2=\sqrt{\dfrac{4}{g}}\rightarrow\:(2)\\ \\ :\implies\sf\:3=\dfrac{1}{2}g{(t_1+t_2+t_3)}^2\\ \\ :\implies\bf\:t_1+t_2+t_3=\sqrt{\dfrac{6}{g}}\rightarrow\:(3)\\ \\ \dag\tt\:subtracting\:(1)\:from\:(2),\:we\:get\\ \\ :\implies\sf\:t_2=\sqrt{\dfrac{4}{g}}-\sqrt{\dfrac{2}{g}}\\ \\ :\implies\bf\:t_2=\sqrt{\dfrac{2}{g}}(\sqrt{2}-1)\rightarrow\:(4)\\ \\ \dag\tt\:subtracting\:(2)\:from\:(3),\:we\:get\\ \\ :\implies\sf\:t_3=\sqrt{\dfrac{6}{g}}-\sqrt{\dfrac{4}{g}}\\ \\ :\implies\bf\:t_3=\sqrt{\dfrac{2}{g}}(\sqrt{3}-\sqrt{2})\rightarrow\:(5)\\ \\ \dag\tt\:From\:(1),(4)\:and\:(5),\:we\:get\\ \\ \longrightarrow\underline{\boxed{\bf{\purple{t_1:t_2:t_3...=1:(\sqrt{2}-1):(\sqrt{3}-\sqrt{2})...}}}}$