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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? |
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Answer» Correct Answer - object, 3.3 s 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 |
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