Question

with what speed should a body be thrown up so that the distance traversed in the 5th and 6th second is the same?

Answer 1

This is only possible if the ball stops moving up post 5th second and starts downward journey at start of 6th second. In this case the distance covered in 5th (between 4 th and 5th sec) and 6th sec (between 5th and 6th) are same, also in 4th and 7th are same, 3rd and 8th too, then 2nd and 9th and finally cover maximum distance amongst all in 1st or 10th sec.

Considering above situation the ball reached maximum height in 5 secs. Atmax height v=0
Therefore using the equation
v=u+at
0 = u - g(5)
u=5g=5x9.8=49 m/s
Thus we get, u=49m/s