A lift accelerates downwards from rest at rate of `2 m//s^(2)`, starting `100 m` above the ground. After 3 sec, an object falls out of the lift. Which will reach the ground first? What is the time interval between their striking the ground?

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