Question

With what speed should a body be thrown upward so that the distance travelled in 5th second and 6th 2nd are equal

Answer 1

As per your question, you have that the body os projected up, taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= t2 then (t1 +t2)= total time of flight.

 

5th second means between t = 4 & t= 5

6th second means  between t=5 & t=6

Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5,

or the time of flight is 10 second or u=50m/s

 

ALTERNATE METHOD

Thus body is at same height at t=4 and t=6

Hence total time of flight is 10 second= 2u/g

thus u = 50m/s

hope this will help.. :-)