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# a coin id dropped in a lift .find the time taken by coin to strike the lift if a.if lift is at restb.lift is going down with v c.lift is going down with constant vd.lift is accelerating with acceleration a upward with u =01(a2(a>g)e.lift is accelerating with acceleration a down with u =01(a2(a>g)

Sharan
20 Points
6 years ago
1. when the lift is at rest the time taken by the coin to strike the lift depends on height as  root 2H/g
2. when lift is going down since their is no comment about the acceleration of the lift therefore the question will be solved as above and T=root 2H/g
3. when lift is going down with constant velocity this implies that acceleration is equal to zero hence again the ans will be T=root 2H/g
4. when the lift is accelerating upwards with acceleration a hence pseudo force will act on the coin opposite to relative motion that is downward therefore expression for T is,
Acceleration of coin = g-a
Therefore time = root 2H / g-a
for a>g T=root 2H/a-g
Acceleration of coin = a + g
Therefore time = root 2H / a+g ​
​         for a>g same as above is done
when the lift is accelerating downwards with acceleration a hence pseudo force will act on the coin opposite to relative motion that is upward therefore expression for T is,