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^{st}` second
S₂ =distance travelled in `\2^{nd}` second
S₃ =distance travelled in `\3^{rd}` second
.
.
.
Sâ‚™ = distance travelled in `\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)

Sâ‚™ = `\frac {1}{2}`g (`\t_n^2` - `\t_{n-1}^2`)

S₁ = `\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}

or 

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


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