Question

the ratio of times taken by freely falling body to cover first metre second metre is​

Answer 1

Answer:

1 : (√2 - 1)

Explanation:

Time taken for first 1 m

t = √(2H/g)

= √(2 × 1/g)

= √(2/g)

Time taken for first 2 m

T = √(2 × 2/g)

= √(4/g)

Time taken for second 1 m is then

T - t = √(4/g) - √(2/g)

Ratio = √(2/g) / [√(4/g) - √(2/g)]

= √2 / (√4 - √2)

= √2 / (√2(√2 - 1))

= 1 : (√2 - 1)

Answer 2

Answer:

The ratio of times taken by freely falling body to cover first metre second metre is​ $\sqrt{1}: \sqrt{2}-\sqrt{1}: \sqrt{3}-\sqrt{2}$

Explanation:

Times taken by freely falling body=  

                                                  $\sqrt{\frac{2H}{g} } \\where H= Height \\g =acceleration due to gravity$

Time taken for first 1 minute, T $=\sqrt{2H / g} =\sqrt{2 / g}$

Time taken for first 2 minute, t $=\sqrt{2 \times 2 / g} =\sqrt{4 / g}$

Time taken for second to 1 minute = $T-t = \sqrt{4 / g}-\sqrt{2 / g}$

Time taken for 1st to 3rd minute $=\sqrt{2 \times \frac{3}{g} } =\sqrt{\frac{6}{g} }$

Time taken for third 1 minute $=\sqrt{6 / g}-\sqrt{4 / g}$

The ratio of times taken by freely falling body to cover first metre, second metre is:

$\ Ratio =\sqrt{2 / g}: \sqrt{4 / g}-\sqrt{2 / g}: \sqrt{6 / g}-\sqrt{4 / g}$