Find the ratio of distance travelled by free falling body in first, second and third second: Unit-3: Motions and Forces (Class 11 -- SLO based Physics Notes)

Find the ratio of distance travelled by the free-falling body in 1st, 2nd and 3rd second: {Ans S₁ : S₂ : S₃ = 1: 3: 5}

Given Data:

t₁ = 1 s
t₂ = 2 s
t₃ = 3 s

To Find:

S₁ : S₂ : S₃ = ?

Solution:

Let
S₁ =distance travelled in

1^{s t}

$\1^{st}$ second
S₂ =distance travelled in

2^{n d}

$\2^{nd}$ second
S₃ =distance travelled in

3^{r d}

$\3^{rd}$ second

Sₙ = distance travelled in

n^{t h}

$\n^{th}$ second

Using 2nd equation of motion, the distance travelled in a specific second for a free-falling object is given by (vi = 0 m/s starting of motion from rest potion)

Sₙ =

\frac{1}{2}

$\frac {1}{2}$ g (

t_{n}^{2}

$\t_n^2$ -

t_{n - 1}^{2}

$\t_{n-1}^2$ )

S₁ =

\frac{1}{2}

$\frac {1}{2}$ g

$\t₁^2$ {for n=1}

S₂ =

$\frac {1}{2}$ g (

$\t₂^2$ -

$\t_1^2$ ) {for n=2}

S₃ =

$\frac {1}{2}$ g (

$\t₃^2$ -

$\t_2^2$ ) {for n=3}

Now the required ratio is

S₁ : S₂ : S₃ =

$\frac {1}{2}$ g

$\1^2$ :

$\frac {1}{2}$ g (

$\2^2$ -

$\1^2$ ) :

$\frac {1}{2}$ g (

$\3^2$ -

$\2^2$ )

S₁ : S₂ : S₃ = (1): (4-1): (9-4) { dividing each ratio by

$\frac {1}{2}$ g}

S₁ : S₂ : S₃ = 1: 3: 5 ------------------Ans

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