Question

For a body that is dropped from a height, the ratio of the distances covered
by the body at the end of 1s, 2 and 3 s, respectively, is
(a) 1:4:9
b 1:2:3
c)1:3:5
(d) 2:6:5
​

Answer 1

Let the body dropped from rest, so after $\sf{n}$ seconds, the distance travelled is,

$\longrightarrow\sf{s_1=0\cdot n+\dfrac{1}{2}g\cdot n^2}$

$\longrightarrow\sf{s_1=\dfrac{gn^2}{2}}$

Since g, acceleration due to gravity, is a constant,

$\longrightarrow\sf{s\propto n^2}$

Hence the ratio of displacements measured from initial point attained after times $\sf{n_1,\ n_2,\ n_3,\dots}$ is,

$\longrightarrow\sf{\underline{\underline{s_1:s_2:s_3:\dots\ =\ (n_1)^2:(n_2)^2:(n_3)^2:\dots}}\quad\quad\dots(1)}$

This is ratio of displacements measured from initial point. The ratio of displacements covered between consecutive times is,

$\longrightarrow\sf{s_1:(s_2-s_1):(s_3-s_2):\dots\ =\ [(n_1)^2-(n_0)^2]:[(n_2)^2-(n_1)^2]:[(n_3)^2-(n_2)^2]:\dots}}}$

Let $\sf{s_i-s_{i-1}=\Delta s_i.}$

$\longrightarrow\sf{\Delta s_1:\Delta s_2:\Delta s_3:\dots\ =\ (n_1)^2:[(n_2-n_1)(n_2+n_1)]:[(n_3-n_2)(n_3+n_2)]:\dots}$

Assume $\sf{n_{i+1}=n_i+1}$ and $\sf{n_0=0.}$ Then we get $\sf{n_1=1.}$ So,

$\longrightarrow\sf{\Delta s_1:\Delta s_2:\Delta s_3:\dots\ =\ (n_1)^2:(2n_1+1):(2n_2+1):\dots}$

In terms of $\sf{n_1,}$

$\longrightarrow\sf{\Delta s_1:\Delta s_2:\Delta s_3:\dots\ =\ (n_1)^2:(2n_1+1):(2n_1+3):\dots}$

$\longrightarrow\sf{\underline{\underline{\Delta s_1:\Delta s_2:\Delta s_3:\dots\ =\ 1:3:5:\dots}}}$

Also, from (1)

$\longrightarrow\sf{\underline{\underline{s_1:s_2:s_3:\dots\ =\ 1:4:9:\dots}}}$

This ratio is known by the name "Galileo's Law of Odd Distances / Numbers".

$\begin{minipage}{10cm}\textit{``The distance covered by a freely falling body is proportional to the square of the time taken for each displacement."}\end{minipage}$

$\begin{minipage}{10cm}\textit{``The distance covered by a freely falling body in same time interval is proportional to consecutive odd numbers."}\end{minipage}$

The case is irrespective of initial velocity; the initial velocity need not be zero.

So the ratio of distances covered at the end of 1s, 2s and 3s respectively, is,

$\longrightarrow\sf{\underline{\underline{s_1:s_2:s_3=1:4:9}}}$