If a particle is thrown vertically upward then its velocity so that it covers same distance in 5th and 6 seconds would be....??

Let u be the initial velocity of the ball. Distance travelled in the nth sec is given by,Sn = u + \frac{1}{2} x a x (2n - 1)Distance covered in 5th second: (put a = -9.8 and n = 5) S5 = u - 44.1Distance covered in the 6th second: (put n = 6, a=-9.8 in equation):S6 = u - 53.9Now, as the body is projected upwards, the distance travelled in two different time intervals will be same only when at one time it is moving upwards and the other time it is coming downwards (as there is symmetry in time and height). So, during 5th second, the object was moving upwards, and during the 6th second, the object is coming downwards. Therefore, the distance S6 will be negative.S6 = -(u - 53.9)Since they are equal, equating S5 and S6 :u - 44.1 = -u + 53.92u = 98u = 49 m/s

Let u be the initial velocity of the ball. Distance travelled in the nth sec is given by,Sn = u + a(n-1/2)Distance covered in 5th second: (put a = -9.8 and n = 5) S5 = u - 44.1Distance covered in the 6th second: (put n = 6, a=-9.8 in equation):S6 = u - 53.9Now, as the body is projected upwards, the distance travelled in two different time intervals will be same only when at one time it is moving upwards and the other time it is coming downwards (as there is symmetry in time and height). So, during 5th second, the object was moving upwards, and during the 6th second, the object is coming downwards. Therefore, the distance S6 will be negative.S6 = -(u - 53.9)Since they are equal, equating S5 and S6 :u - 44.1 = -u + 53.92u = 98u = 49 m/s