A coin is released inside a lift at a height of $2$ m from the floor of the lift. The height of the lift is $10$ m. The lift is moving with an acceleration of $9$ $m / s^{2}$ downwards. The time after which the coin will strike the lift is :

4 s

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2 s

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\frac{4}{\sqrt{21}} s

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\frac{2}{\sqrt{11}} s

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Solution

The correct option is B 2 s
$I n i t i a l v e l o c i t y r e l a t i v e t o l i f t \Rightarrow 0 m / s R e l a t i v e a c c e l e r a t i o n \Rightarrow 10 - 9 = 1 m / s^{2} ∴ s = \frac{1}{2} a_{r e l} t^{2} \Rightarrow 2 = \frac{1}{2} \times 1 \times t^{2} \Rightarrow t^{2} = 4 \Rightarrow t = 2 s e c o n d s .$

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A coin is released inside a lift at a height of 2 m from the floor of the lift. The height of the lift is 10 m. The lift is moving with an acceleration of 9 m/s2 downwards. The time after which the coin will strike the lift is :

The correct option is B 2 sInitialvelocityrelativetolift⇒0m/sRelativeacceleration⇒10−9=1m/s2∴s=12arelt2⇒2=12×1×t2⇒t2=4⇒t=2seconds.

A coin is released inside a lift at a height of $2$ m from the floor of the lift. The height of the lift is $10$ m. The lift is moving with an acceleration of $9$ $m / s^{2}$ downwards. The time after which the coin will strike the lift is :