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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.
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
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
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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