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Let *u* be the initial velocity of the ball. Distance travelled in the *n ^{th }sec *is given by,

S_{n }= u + x a x (2n - 1)

Distance covered in 5th second: (put a = -9.8 and n = 5)

S_{5 }= u - 44.1

Distance covered in the 6th second: (put n = 6, a=-9.8 in equation):

S_{6 }= u - 53.9

Now, 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 S_{6 }will be negative.

S_{6 }= -(u - 53.9)

Since they are equal, equating S_{5} and S_{6} :

u - 44.1 = -u + 53.9

2u = 98

**u = 49 m/s**

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