Kinematics- Physics

With what speed should a body be thrown upwards so that the distances traversed in the 5th second and 6th second are equal?

18-Nov-2015 2:44 PM

Answers (2)


Let u be the initial velocity of the ball. Distance travelled in the nth sec is given by,

S= u + \frac{1}{2} x a x (2n - 1)

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

S= u - 44.1

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

S= 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 Swill be negative.

S=  -(u - 53.9)

Since they are equal, equating S5 and S6 :

u - 44.1 = -u + 53.9

2u = 98

u = 49 m/s

18-11-2015 15:05


Thanks a lot!!! Your explanation is really clear and helpful :)

18-11-2015 17:44

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