A particle is thrown up vertically with a speed `v_(1)` in air. It takes time `t_(1)` in upwards journey and `t_(2) (gt t_(1))` in the downward journey and returns to the starting point with a speed `v_(2)`. Then:

Answer

object, 3.3 s

Explanation

After 3 sec distance covered `=1//2xx2xx9=9m` velocity of life`=2xx3=6 m//s darr :. U_(p)=6m//s darr, a=g darr` height `=(100-9)=91 m`
`:.` Time to reach the ground
`=91=6t+1/2xxgxxt^(2) t=3.7` sec
Total time taken by object to reach the ground
`=3+3.7=6.7` sec
Time to reach on the ground by lift
`=1/2xx2xxt^(2)=100 rArr t=10` sec
So interval `=10-6.7=3.3`sec