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An elevator car whose floor to ceiling d...

An elevator car whose floor to ceiling distance is equal to `2.7 m` starts ascending with constant acceleration `1.2 m//s^(2)`, 2 sec after the start a bolt begins falling from the ceiling of the car. Answer the following question `(g=9.8 m//s^(2))`
The bolt's free fall time is

A

`sqrt((2.7)/(9.8))s`

B

`sqrt((5.4)/(9.8))s`

C

`sqrt((5.4)/(8.6))s`

D

`sqrt((5.4)/(11))s`

Text Solution

Verified by Experts

The correct Answer is:
D
Relative to lift,
`u_(r )=0, a_(r )=(9.8+1.2)=11 ms^(-2)s_(r )=(1)/(2)a_(r )t^(2)`
`therefore 2.7=(1)/(2)xx11xxt^(2) rArr t = sqrt((5.4)/(11))s`
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