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`        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.`
7 years ago

```							2*(2m)/(7.8m/s2)=(answer)^2
```
7 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
```
7 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
```
7 years ago
```							.5sec
```
7 years ago
```							1/2^1/2
```
7 years ago
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