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


Given Data:


t₁ 
= 1 s
t
 = 2 s
t
 = 3 s


To Find:

S₁ S₂ S = ?


Solution: 


Let
S₁ 
=distance travelled in 
1st second
S
 =distance travelled in 2nd second
S
 =distance travelled in 3rd second
.
.
.
Sₙdistance travelled in nth 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 12g (t2n - t2n-1)

S₁ 12g t2   {for n=1}
S = 12g (t2 - t21)  {for n=2}
S 12g (t2 - t22) {for n=3}

Now the required ratio is
S₁ : S₂ : S₃ = 12 g 12 : 12 g (22 - 12) : 12 (32 - 22)

S₁ S₂ S = (1): (4-1): (9-4)      { dividing each ratio by 12 g}

or 

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



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