a lift is moving with uniform downward acceleration of 2m/s2 . a ball is dropped from a height 2m from yhe floor of lift. find time after which ball strike the floor.

2*(2m)/(7.8m/s2)=(answer)^2

Relative acceleration= (g-a)= (10-2)m/s= 8m/s 

hence H=ut+1/2 a t^2

putting u=0, a=8, h=2 we get

t= 0.7 s

this question can be solved by the concept of relative motion

acceleration of ball relative to lift = acceleration of ball relative to ground - acc. of lift with respect to ground

                                               = -10j - (- 2j)  {note: we have used j component and it is in downward direction therefore it is                                                                       negative}

                                               = -8j

now using the relation s = ut + 1/2 a t^2

     u=0    s=2m    a = +8m/s^2  (the sign here used is positive because the acceleration -8j will help in moving the ball                                                           downward)

2 = 1/2 (8) t^2

t = 2 ^1/2 (root 2) seconds 

.5sec

1/2^1/2