# Question #65d5d

Jul 15, 2016

Kinematic equation for a free falling body under gravity is given by
$s = u t + \frac{1}{2} g {t}^{2}$
Let ${s}_{1}$ be the distance traveled in time ${t}_{1}$
similarly ${s}_{2} \mathmr{and} {s}_{3}$ be distances traveled in time ${t}_{2} \mathmr{and} {t}_{3}$ respectively. As per original conditions
${s}_{1} = \frac{1}{2} g {t}_{1}^{2}$
${s}_{2} = \frac{1}{2} g {t}_{2}^{2}$
${s}_{3} = \frac{1}{2} g {t}_{3}^{2}$
Distance traveled in time interval ${t}_{2} - {t}_{1}$ $= {s}_{2} - {s}_{1}$
$\implies {s}_{2} - {s}_{1} = \frac{1}{2} g {t}_{2}^{2} - \frac{1}{2} g {t}_{1}^{2}$
$\implies {s}_{2} - {s}_{1} = \frac{1}{2} g \left({t}_{2}^{2} - {t}_{1}^{2}\right)$ ......(1)
Similarly ${s}_{3} - {s}_{2} = \frac{1}{2} g \left({t}_{3}^{2} - {t}_{2}^{2}\right)$ ......(2)

Ratio of the distances traveled in successive intervals can be found by dividing (1) by (2)
$\frac{{s}_{2} - {s}_{1}}{{s}_{3} - {s}_{2}} = \frac{\frac{1}{2} g \left({t}_{2}^{2} - {t}_{1}^{2}\right)}{\frac{1}{2} g \left({t}_{3}^{2} - {t}_{2}^{2}\right)}$
$\frac{{s}_{2} - {s}_{1}}{{s}_{3} - {s}_{2}} = \frac{{t}_{2}^{2} - {t}_{1}^{2}}{{t}_{3}^{2} - {t}_{2}^{2}}$

Jul 15, 2016

$\text{Ratios is shown in figure}$

#### Explanation:

$\text{example: ratio for t=20 and t=21}$

$\frac{\left(2 \cdot 20 - 1\right) \cdot x}{\left(2 \cdot 21 - 1\right) \cdot x} = \frac{39 \cdot x}{41 \cdot x} = \frac{39}{41}$