Lift A is going up with velocity 20 m/s and acceleration a equal to $8 m / s^{2}$ in the downward direction. While lift B is going down with 10 m/s and acceleration of +2 $m / s^{2}$ in upward direction. At this moment a bolt in the ceiling of lift A gets loose and falls. Height of the lift is 16 m. Find the time taken by bolt to hit the floor of A.

2 sec

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4 sec

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Solution

The correct option is C
4 sec

$v_{b / A} = 0, a_{b / A} = - g - (- a_{A}) = - 2 m / s^{2}$
$s_{b / A} = - 16 m$
$T = \sqrt{\frac{2 S_{b / A}}{a_{b / A}}} = \sqrt{(\frac{2 \times 16}{2})} = 4 s$

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Similar questions

Q. Lift A is going up with velocity 20m/s and acceleration as

8 m / s^{2}

in the downward direction. While lift B is going down with 10m/s and acceleration of

+ 2 m / s^{2}

in upward direction. At this moment a bolt in the ceiling of lift A gets loose and falls. Height of the lift is 16 m.

What is the velocity of bolt relative to lift B when it hits the floor of A.?

Lift A is going up with velocity 20 m/s and acceleration as $8 m / s^{2}$ in the downward direction. While lift B is going down with 10 m/s and acceleration of $+ 2 m / s^{2}$ in upward direction. At this moment a bolt in the ceiling of lift A gets loose and falls. Height of the lift is 16 m.

What is the velocity of bolt relative to lift B when it hits the floor of A.?

Lift A is going up with velocity 20m/s and acceleration of $8 m / s^{2}$ in the downward direction. While lift B is going down with 10m/s and acceleration of $+ 2 m / s^{2}$ in upward direction. At this moment a bolt in the ceiling of lift A gets loose and falls. Height of the lift is 16 m.Find the displacement of bolt relative to B when it hits the floor of A.

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The correct option is C 4 sec vb/A=0, ab/A=−g−(−aA)=−2m/s2 sb/A=−16m T=√2Sb/Aab/A=√(2×162)=4s

The correct option is C
4 sec

$v_{b / A} = 0, a_{b / A} = - g - (- a_{A}) = - 2 m / s^{2}$
$s_{b / A} = - 16 m$
$T = \sqrt{\frac{2 S_{b / A}}{a_{b / A}}} = \sqrt{(\frac{2 \times 16}{2})} = 4 s$