Question

A particle is released from rest from a tower of height 3h. The ratio of the intervals of time to cover three equal heights h is

• A. $t_{1}:t_{2}:t_{3} = 3:2:1$
• B. $t_{1}:t_{2}:t_{3} = 1:\left( \sqrt{2} - 1 \right):\left( \sqrt{3}:2 \right)$
• C. $t_{1}:t_{2}:t_{3} = \sqrt{3}:\sqrt{2}:1$
• D. $t_{1}:t_{2}:t_{3} = 1:\left( \sqrt{2} - 1 \right):\left( \sqrt{3} - \sqrt{2} \right)$

Answer 1

Answer: (D)

$t_{1}:t_{2}:t_{3} = 1:\left( \sqrt{2} - 1 \right):\left( \sqrt{3} - \sqrt{2} \right)$

Answer 2

Let $t_{1},t_{2},t_{3}$be the timings for three successive equal heights h covered during the free fall of the particle.

Then

$h = \frac{1}{2}gt_{1}^{2}$ ($\because u = 0$)

Or $t_{1} = \sqrt{\frac{2h}{g}}$ ….. (i)

$2h = \frac{1}{2}g(t_{1} + t_{2})^{2}$

Or $t_{1} + t_{2} = \sqrt{\frac{4h}{g}}$ …… (ii)

$3h = \frac{1}{2}g(t_{1} + t_{2} + t_{3})^{2}$

Or $t_{1} + t_{2} + t_{3} = \sqrt{\frac{6h}{g}}$ ……. (iii)

Subtracting (i) from (ii), we get

$t_{2} = \sqrt{\frac{4h}{g}} - \sqrt{\frac{2h}{g}} = \sqrt{\frac{2h}{g}}(\sqrt{2} - 1)$ ……. (iv)

Subtracting (ii) form (iii), we get

$t_{3} = \sqrt{\frac{6h}{g}} - \sqrt{\frac{4h}{g}} = \sqrt{\frac{2h}{g}}(\sqrt{3} - \sqrt{2})$ ……. (v)

From (i),(iv) and (v),we get

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