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A particle is thrown vertically upward from ground at 55 m/s. Distance travelled by the body in 6th second is (g = 10 m/s^2) 1) 2.5 m 2) 5 m 3) 10 m 4) 1.25 m
It means distance travelled between t=5s to t=6s.First of all we have to check if the body reaches its maximum height before 6s or not.At maximum height velocity v=0.v=u-gt=55-10tOr,0=55-10tt=5.5sSo after 5.5 second the body starts falling back.At t=5s the velocity of the body isv'=u-gt=55-10×5=5m/sFrom t=5 to t=5.5s the body travels upward.The distance coveredh=v't-gt2/2Here t=5.5-5=0.5s.h=5×0.5-10×0.52/2=1.25 mFrom t=5.5 to t=6s the body falls downward.at t=5.5 velocity is zero.so distance covered in this 0.5 s ish=ut+gt2/2=0+10×0.52/2=1.25mAdding these two distance covered between 5s to 6s is 2.5If it helps please approve
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