A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of $1\, m$ each will then be
A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of $1\, m$ each will then be
All equal, being equal to $\sqrt {2/g} $ second
In the ratio of the square roots of the integers $1, 2, 3.....$
In the ratio of the difference in the square roots of the integers i.e. $\sqrt 1 ,\,(\sqrt 2 - \sqrt 1 ),\,(\sqrt 3 - \sqrt 2 ),\,(\sqrt 4 - \sqrt 3 )$....
In the ratio of the reciprocal of the square roots of the integers i.e.,. $\frac{1}{{\sqrt 1 }},\,\frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 4 }}$
A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distances of $1\, m$ each will then be
$h = ut + \frac{1}{2}g{t^2} \Rightarrow 1 = 0 \times {t_1} + \frac{1}{2}gt_1^2 \Rightarrow {t_1} = \sqrt {2/g} $
Velocity after travelling $1\,m $ distance
${v^2} = {u^2} + 2gh \Rightarrow {v^2} = {(0)^2} + 2g \times 1 \Rightarrow v = \sqrt {2g} $
For second $1$ meter distance $1 = \sqrt {2g} \times {t_2} + \frac{1}{2}gt_2^2 \Rightarrow gt_2^2 + 2\sqrt {2g} {t_2} - 2 = 0$
${t_2} = \frac{{ - 2\sqrt {2g} \pm \sqrt {8g + 8g} }}{{2g}} = \frac{{ - \sqrt 2 \pm 2}}{{\sqrt g }}$
Taking $+ve$ sign ${t_2} = (2 - \sqrt 2 )/\sqrt g $
$\therefore \frac{{{t_1}}}{{{t_2}}} = \frac{{\sqrt {2/g} }}{{(2 - \sqrt 2 )/\sqrt g }} = \frac{1}{{\sqrt 2 - 1}}$ and so on.