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Question

the displacement and the distance covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.

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Solution

At the moment the bolt contact with the elevator, it has already acquired the velocity equal to elevator given by:
ν0=(1.2)(2)=2.4 m/s
In the reference frame attached with the elevator shaft (ground) and pointing the y axis upward, we have for the displacement of the bolt,
Δy=ν0y+12wyt2
=ν0+12(g)t2
or, Δy=(2.4)(0.7)+12(9.8)(0.7)2=0.7 m
Hence the bolt comes down or displace downward relative to the point, when it loses contact with the elevator by the amount 0.7 m (Fig).
Obviously the total distance covered by the bolt during its free fall time
s=|Δy|+2(ν02g)=0.7m+(2.4)2(9.8)m=1.3 m

1793116_1851194_ans_732f3152e5c54212b3b8e29c6d341873.png

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