# 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.

Chetan Mandayam Nayakar
312 Points
11 years ago

Pavel Banik
20 Points
11 years ago

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

James Macmillan
20 Points
11 years ago

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

shashank gattani
18 Points
11 years ago

.5sec

James Macmillan
20 Points
11 years ago

1/2^1/2