An elevator ascends with an upward acceleration of $0.2 m / s^{2}$ . At the instant when its upward speed is $2 m / s$ , a loose bolt $5 m$ high from the floor drops from the ceiling of the elevator. The time taken by the bolt to strike the floor and the distance it has fallen are (Take $g = 9.8 m / s^{2}$ )

1 s

1.9 m

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1 s

2.9 m

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1 s

4.9 m

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1 s

3.9 m

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Solution

The correct option is C $1 s$ , $4.9 m$
$g = 9.8 a = 0.2 a_{r e l} = 10 N o w, s = \frac{1}{2} a_{r e l} t^{2} t = \sqrt{\frac{2 \times 5}{10}} = 1 s e c o n d . D i s t a n c e c o v e r e d b y e l e v a t o r = \frac{1}{2} \times 0.2 \times 1 ∴ D i s p l a c e m e n t o f b o l t = 5 - 0.1 = 4.9 m e t e r s .$

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An elevator ascends with an upward acceleration of 2 $m s^{- 2}$ . At the instant its upward speed is 2.5 $m s^{- 1}$ , a loose bolt is dropped from the ceiling of the elevator 3m from the floor. ( $g = 10 m s^{- 2}$ ). The time of flight of the bolt from the ceiling to floor approximately is