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

(1) All equal, being equal to $\sqrt{2/g}$ second

(2) In the ratio of the square roots of the integers 1, 2, 3.....

(3) In the ratio of the difference in the square roots of the integers i.e. $\sqrt{1},\text{\hspace{0.17em}}\left(\sqrt{2}-\sqrt{1}\right),\text{\hspace{0.17em}}\left(\sqrt{3}-\sqrt{2}\right),\text{\hspace{0.17em}}\left(\sqrt{4}-\sqrt{3}\right)$ ....

(4) In the ratio of the reciprocal of the square roots of the integers i.e.,. $\frac{1}{\sqrt{1}},\text{\hspace{0.17em}}\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\text{\hspace{0.17em}}\frac{1}{\sqrt{4}}$

(3) $h=ut+\frac{1}{2}g{t}^{2}⇒1=0×{t}_{1}+\frac{1}{2}g{t}_{1}^{2}⇒{t}_{1}=\sqrt{2/g}$

Velocity after travelling 1m distance

${v}^{2}={u}^{2}+2gh⇒{v}^{2}={\left(0\right)}^{2}+2g×1⇒v=\sqrt{2g}$

For second 1 meter distance

$1=\sqrt{2g}×{t}_{2}+\frac{1}{2}g{t}_{2}^{2}⇒g{t}_{2}^{2}+2\sqrt{2g}{t}_{2}-2=0$

${t}_{2}=\frac{-2\sqrt{2g}±\sqrt{8g+8g}}{2g}=\frac{-\sqrt{2}±2}{\sqrt{g}}$

Taking +ve sign ${t}_{2}=\left(2-\sqrt{2}\right)/\sqrt{g}$

$\frac{{t}_{1}}{{t}_{2}}=\frac{\sqrt{2/g}}{\left(2-\sqrt{2}\right)/\sqrt{g}}=\frac{1}{\sqrt{2}-1}$ and so on.

Difficulty Level:

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