Question

A ball is dropped from a height, find the ratio of displacement travelled in first, second and third seconds?

Answer 1

Given:

A ball is dropped from a height.

To find:

The ratio of displacement travelled in first, second and third seconds?

Calculation:

We will use displacement equation for n_(th) second:

• $s_(n^{th})=u + \dfrac{1}{2}a(2n-1)$

In the first second :

$\sf \: s_{1} = u + \dfrac{1}{2} g \{2(1) - 1 \}$

$\sf \implies\: s_{1} = 0+ \dfrac{1}{2} g \{2(1) - 1 \}$

$\sf \implies\: s_{1} =5 \: m$

In the 2nd second:

$\sf \: s_{2} = u + \dfrac{1}{2} g \{2(2) - 1 \}$

$\sf \implies\: s_{2} = 0+ \dfrac{1}{2} g \{2(2) - 1 \}$

$\sf \implies\: s_{2} = 15 \: m$

In the 3rd second:

$\sf \: s_{3} = u + \dfrac{1}{2} g \{2(3) - 1 \}$

$\sf \implies\: s_{3} = 0+ \dfrac{1}{2} g \{2(3) - 1 \}$

$\sf \implies\: s_{3} = 25\: m$

So, ratio is :

$\boxed{ \bf s_{1} : s_{2} : s_{3} = 1 : 3 : 5}$