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A body falling from rest covers distances s1 and s2 in first and second seconds respectively. Calculate the ratio s1/s2.

A body falling from rest covers distances s1 and s2 in first and second seconds respectively. Calculate the ratio s1/s2.

Grade:10

3 Answers

vikas askiitian expert
509 Points
10 years ago

S = ut + at2/2                            (a=g & u=0 & t=1)       

(g is accleration due to gravity)

S1 = g/2                                        (distance covered in 1 sec)

distance covered in 2 sec = ut+at2/2                              (u=0 , a=g & t=2)

                                    =2g

distance covered in 2nd second =S2= 2g-g/2=3g/2

ratio =S1/S2 = 1/3

nikhil arora
54 Points
10 years ago

distance covered in 1st second=

s1=ut+(1/2)at(sqr)

s1=(1/2)g

distance covered in 2nd second=distance covered in 2 seconds - distance covered in 1 second

s2=(1/2)gx4 - (1/2)gx1

s2=(3/2)g

 

s1/s2=1:3

Fawz Naim
37 Points
10 years ago

the formula for the distance covered by a body falling freely under gravity is=u+a(2t-1)/2

u is the initial velocity, a is the acc., t is  second

in first second s1=u+g(2t-1)/2

u=0, t=1

s1=g/2

similarly

s2=0+g(2*2-1)/2

s2=3g/2

ratio of s1/s2=g/2/3g/2

=1/3

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