Two bodies are thrown upwards with same initial velocity 40 ms-1  with time gap 4 sec. After what time from projection of first body they will meet  (g  = 10 m/s2)

Displacement covered by first object in t secs⁡,  s1=ut−12gt2=40t−1210×t2  Displacement covered by second object in (t−4)secs⁡, s2=40(t−4)−1210×(t−4)2  Since they meet s1=s2  We get t=6 secs