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body travels half of its total distance in last second of its fall from rest, then time of fall is (Assume g = 9.8 ms –2 )
lets say total time in fall is t so S(t)=s =1/2gt2and s(t-1)=s/2 =1/2(g-1)2devideing both2={t/(t-1)}2t/(t-1)=sqrt(2)t=
Last second of fall means t`=t-1Displacement in nth second =u+1/2*(2t`-1)Total displacement =ut+ 1/2*at^2Displacement of body in nth second=0+1/2*9.8(2(t-1)-1)Let total displacement be S,S/2=4.9(2t-3)...........1Total displacement,S=1/2*9 .8*t^2S=4.9t^2...........2Putting valve of S from 2 into 1,4.9t^2=9.8*(2t-3)4.9t^2=19.6t-29.4Dividing whole by 4.9,t^2=4t-6t^2-4t+6=0Applying sridharacharya formula we get,t=2+√2seconds
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