Question

THE RATIO OF TIMES TAKEN BY A FREELY
FALLING BODY TO COVER FIRST METER
SECOND METER AND THIRD METER​

Answer 1

SoLuTiOn :

➡ Given:

✏ A case of free falling body.

➡ To Find:

✏ The ratio of time taken to cover first meter, second meter and third meter.

➡ Formula:

✏ As per second equation of kinematics formula of distance covered by body is given by

$\star \: \underline{ \boxed{ \bold{ \rm{ \pink{h = ut + \frac{1}{2} a {t}^{2} }}}}} \: \star$

➡ Calculation:

✏ Let t1, t2, t3 be the timings for three successive equal heights h covered during the free fall of the particle.

Then,

$\mapsto \rm \: h = \frac{1}{2} g {t}^{2} \: or \: t_1 = \sqrt{ \frac{2h}{g} } .....(1) \: \: ( \because{u} = 0) \\ \\ \mapsto \rm \: 2h = \frac{1}{2} g {(t_1 + t_2)}^{2} \: or \: t_1 + t_2 = \sqrt{ \frac{4h}{g} } .....(2) \\ \\ \mapsto \rm \: 3h = \frac{1}{2} g {(t_1 + t_2 + t_3)}^{2} \: or \: t_1 + t_2 + t_3 = \sqrt{ \frac{6h}{g} } .....(3)$

✏ Substracting 1 from 2, we get

$\mapsto \rm \: t_2 = \sqrt{ \frac{4h}{g} } - \sqrt{ \frac{2h}{g} } = \sqrt{ \frac{2h}{g} } ( \sqrt{2} - 1).....(4)$

✏ Substracting 2 from 3, we get

$\mapsto \rm \: t_3 = \sqrt{ \frac{6h}{g} } - \sqrt{ \frac{4h}{g} } = \sqrt{ \frac{2h}{g} } ( \sqrt{3} - \sqrt{2} ).....(5)$

✏ From 1, 4 and 5, we get

$\orange{\bigstar}\underline{ \boxed{ \bold{ \rm{ \purple{t_1 : t_2 : t_3 = 1 :( \sqrt{2} - 1) : ( \sqrt{3} - \sqrt{2} ) }}}}}\orange{\bigstar}$