Question

With what speed should a body with thrown upward so that the distance Traversed in 5th second and 6 that second are equal

Answer 1

This is only possible when the body reaches maximum height at 5th sec and starts downfall thereafter.
Why??
Because energy is conserved. At every second, body moves up, velocity decreases until zero. When it moves down, velocity increases. So, at any other instant, other than at maximum height will diatances covered be equal.

So, 1st second means 0s to 1s.

So, 5th second means 4th second to 5th second., it moves up.

At 5th second, it is at top, ie, v=0.

6th second means from 5th to 6th second, body moves down.

Now, that means body must reach zero velocity at t=5 .

So,
At t=5
v=0
a=g=9.8m/s
V= u-gt for upward motion.
0= u - g×5
u= 5g.
Assuming g =9.8,
u = 5×9.8 = 49m/s .

So, body must be thrown up at 49m/s, so that it covers equal distance at 4th and 5th second.

You can verify using other equations for constant displacement at given times.