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Jayati Dinesh Bharadwaj Grade: 10
        

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

6 years ago

Answers : (3)

vikas askiitian expert
510 Points
										

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

6 years ago
nikhil arora
54 Points
										

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

6 years ago
Fawz Naim
37 Points
										

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

6 years ago
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