An elevator car whose floor to ceiling distance is equal to `2.7m` starts ascending with constant acceleration `1.2 m//s^2.` 2 s after the start, a bolt begins falling from the ceiling of the car. Find
(a)the time after which bolt hits the floor of the elevator.
(b)the net displacement and distance travelled by the bolt, with respect to earth. (Take `g=9.8 m//s^2)`

Answer

(a) `0.7s`; (b) `0.7` and `1.3m` respectively.

Explanation

(a) One good way to solve the problem is to work in the elevator's frame having the observer at its bottom(figure).
Let us denote the separation between floor and celing by `h=2*7m.` and the acceleration of the elevator by `w=1*2m//s^2`
From the kinematical formula
`y=y_0+nu_(0y)t+1/2w_yt^2` (1)
Here `y=0`, `y_0=+h`, `nu_(0y)=0`
and `w_y=w_("bolt"(y))-w_("ele"(y))`
`=(-g)-(w)=-(g+w)`
So, `0=h+1/2{-(g+w)}t^2`
or, `t=sqrt((2h)/(g+w))=0*7s`.

(b) At the moment the bolt loses contract with the elevator, it has already aquired the velocity equal to elevator given by:
`v_0=(1*2)(2)=2*4m//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,
`Deltay=nu_(0y)t+1/2w_yt^2`
`=v_0t+1/2(-g)t^2`
or, `Deltay=(2*4)(0*7)+1/2(-9*8)(0*7)^2=-0*7m.`

Hence the bolt comes down or displaces downward relative to the point, when it loses contact with the elevator by the amount `0*7m`(figure).
Obviously the total distance covered by the bolt during its free fall time
`s=|Deltay|+2((v_0^2)/(2g))=0*7m+((2*4)^2)/((9*8))m=1*3m`.