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)`)

Options

A. 0.3 second
B. 0.2 second
C. 0.16 second
D. 0.4 second

Answer

D

Explanation

Time taken by the ball to reach the floor of the lift is
`t = sqrt((2h)/(a + g))`
Here, h = 1.25 m, g = 1`0 ms^(-2)`, a = `5 ms^(-2)`
`t = sqrt((2 xx 1.25 m)/((5 + 10) m s^(-2)))` = 0.4 s