Question

A lift is moving down with acceleration of 3ms^-2.A ball is released 1.7m above the floor.Assuming that g=9.8ms^-2,how long will the ball take to hit the floor

Answer 1

@Abhisar

Answer 2

Do u know the answer, i mean do u have an answer key ?

Answer 3

Maybe,I need to use
s=ut+1/2at^2

Answer 4

yeah i know that, just wanted to confirm my answers. Use the same equation and take a=9.8

Answer 5

The given answer is 0.71s

Answer 6

is ur question complete ?

Answer 7

yes

Answer 8

Something is incomplete, when substituting the value
Nt sure, whether this is correct
I take 9.8-3=a
I can get the correct answer

Answer 9

a will be 9.8+3

Answer 10

I tried that earlier
I'm getting 0.27 as my final answer

Answer 11

oh yes i got it !

Answer 12

You will have to use the concept of relative motion.

Answer 13

Since both the lift and ball are in motion that too with different accelerations. You will have to apply relative motion concept. Acceleration of ball = 9.8 (since it is not in contact with lift), acceleration of lift = 3. So acceleration of ball relative to lift will be 9.8-3= 6.8

Answer 14

now take a=6.8 and use the second equation to find the value of t

Answer 15

Okay
Understood
t^2=0.5
t=0.71
Thanks @Abhisar

Answer 16

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