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

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Susmita · Answer

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=55-10t Or,0=55-10t t=5.5s So after 5.5 second the body starts falling back. At t=5s the velocity of the body is v'=u-gt=55-10×5=5m/s From t=5 to t=5.5s the body travels upward.The distance covered h=v't-gt 2 /2 Here t=5.5-5=0.5s. h=5×0.5-10×0.5 2 /2=1.25 m From 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 is h=ut+gt 2 /2=0+10×0.5 2 /2=1.25m Adding these two distance covered between 5s to 6s is 2.5 If it helps please approve

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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

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

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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

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 m2) 5 m3) 10 m4) 1.25 m

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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