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Physics 18 Online
OpenStudy (anonymous):

an elevator is going down with a constant acceleration. A coin is dropped from a point 1.5m above the elevator floor takes one second to reach the floor. The magnitude of the acceleration of elevator is??? (g=10m/s2)

OpenStudy (experimentx):

s = ut +(g-a)t^2, u =0 1.5 = 10-a, a = 8.5m/s^2

OpenStudy (anonymous):

hey ur answer is wrong option are (a) 3.6 (b) 5 (c)7.2 (d)6.4

OpenStudy (experimentx):

sorry for wrong formula s = ut +1/2 *(g-a)t^2,

OpenStudy (anonymous):

then u=?????

OpenStudy (experimentx):

0

OpenStudy (anonymous):

how??

OpenStudy (experimentx):

|dw:1336272871420:dw| Whenever body moves from rest, the initial velocity is 0

OpenStudy (anonymous):

in this question it is not mentioned it is on rest.......

OpenStudy (experimentx):

i don't think you can find if it is not given any value.

OpenStudy (anonymous):

the the answer is 6.4 but i dont know how to get it

OpenStudy (anonymous):

sorry my question was wrong

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