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In a lift moving up with an acceleration...

In a lift moving up with an acceleration of `5 m s^(-2)`, a ball is dropped from a height of 1.25 m. The time taken by the ball to reach the floor of the lift is ……(nearly) (`g = 10 ms^(-2)`)

A

0.3 second

B

0.2 second

C

0.16 second

D

0.4 second

Text Solution

Verified by Experts

The correct Answer is:
d
Since life is ascending upward with acceleration a. So when the ball is dropped then it well fall with an acceleration of g+a i.e 10+5 =`15m//s^(2)`
`:.` Time of descnt is `T=sqrt((2h)/(g+a))=sqrt((2xx1.5)/(15))=0.4s`
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