Question

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. All equal, being equal to $\sqrt{\frac{2}{g}}$ second
• B. In the ratio of the square roots of the integers 1, 2, 3.....
• C. In the ratio of the difference in the square roots of the integers i.e. $\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3})...$
• D. In the ratio of the reciprocal of the square roots of the integers i.e., $\frac{1}{\sqrt{1}},\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{4}}$

Answer 1

The correct option is C In the ratio of the difference in the square roots of the integers i.e. $\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3})...$

$h=ut+\frac{1}{2}g{t}^2\Rightarrow 1=0\times t_1+\frac{1}{2}g{t}_1^2\Rightarrow t_1=\sqrt{\frac{2}{g}}$
Velocity after travelling 1m distance
$v^2=u^2+2gh\Rightarrow v^2=(0{)}^2+2g\times 1\Rightarrow v=\sqrt{2g}$
For second 1 meter distance $1=\sqrt{2g}\times t_2+\frac{1}{2}g{t}_2^2\Rightarrow g{t}_2^2+2\sqrt{2g{t}_2}-2=0$
$t_2=\frac{-2\sqrt{2g}\pm \sqrt{8g+8g}}{2g}=\frac{-\sqrt{2}\pm 2}{\sqrt{g}}$
Taking +ve sign $t_2=\frac{2-\sqrt{2}}{\sqrt{g}}$
$\therefore$ $\frac{t_1}{t_2}=\frac{\sqrt{\frac{2}{g}}}{\frac{(2-\sqrt{2})}{\sqrt{g}}}=\frac{1}{\sqrt{2}-1}$ and so on