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15 Apr 2012 17:10:10 IST

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A lift is descending with uniform acceration.To measure the acceleration,a person in the lift drops a coin at the moment lift starts.The coin is 6 ft above the floor of the lift at time it is dropped.The person observes that the coin strikes the floor in 1 second.Calculate the acceleration of lift.

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ashwini

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15 Apr 2012 17:55:46 IST

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this one is too easy u better clear ur fundamentals b4 posting on forum blindly...............2 ft /sec2 upwards

akshara

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15 Apr 2012 17:59:58 IST

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its not 2ft/s2

ashwini

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15 Apr 2012 18:01:30 IST

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OHHHHHH COMMON ..........DO U STUDY FROM SOME LOW LEVEL BOOK THAT HAS WRONG ANSWERS

akshara

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15 Apr 2012 18:03:56 IST

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its from H C VERMA

harsh

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15 Apr 2012 18:23:41 IST

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8m/s² downward

Tarun Garg

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15 Apr 2012 18:29:09 IST

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answer is 31 ft/s2

Tarun Garg

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15 Apr 2012 18:37:03 IST

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No its 20 ft/s2

Tarun Garg

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15 Apr 2012 18:49:40 IST

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No my final is 20

akshara

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15 Apr 2012 18:50:02 IST

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S.its 20ft/s2.can u please give me the method to solve this

Rhrishikesh Sane

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15 Apr 2012 18:51:19 IST

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@ashwini .............cant u just shut up............ Tarun is right...........ugave the rong answer only.............at least he tried think about it again and got his mistake and accepted it.........and u wer givin excuses that akshara uses some low level book...............so cheap..............and dont use abusive language as u did in case of sunita banergy............udidnt know the correct reason for cold drink so u answered like that............

Tarun Garg

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15 Apr 2012 19:03:12 IST

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U have to just know that g=32.34 ft/s2 and net acc.=g-acc. Of lift

Tarun Garg

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15 Apr 2012 19:03:59 IST

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U have to just know that g=32.34 ft/s2 and net acc.=g-acc. Of lift

Tarun Garg

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15 Apr 2012 19:04:37 IST

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Hit lyk if u got it

harsh

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15 Apr 2012 19:04:52 IST

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here u can apply the concept of relative velocity , jus take the lift as "a" then the acceleration of coin wrt lift is -g-a , so in other way u can take lift at rest and coin is coming downward with acceleration g+a , and the distance between lift and the coin is 6ft as given, the time taken by coin to reach the floor is 1sec, and initial velocity of coin i.e u=0 , so just apply s=ut +1/2(atsquare).where a is -a-g, t is 1sec and s is -6 ft. i hope it will help.

harsh

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15 Apr 2012 19:06:54 IST

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here take the acceleration of lift as "a" as i mentioned in first line.

akshara

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15 Apr 2012 19:27:47 IST

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thnx tarun n harsh.i got it

harsh

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15 Apr 2012 19:50:22 IST

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MY PLEASURE , U CAN ASK ANY TIME, TRY MORE DIFFICULT QUESTIONS ,AND IF U WILL NOT BE ABLE TO SOLVE IT POST IT HERE , WE WILL ALSO TRY

hrishikesh sane

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15 Apr 2012 19:52:40 IST

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@HARSH..........STOP FLATTERING...EVERYONE KNOWS HOW MUCH KNACK U HV WHEN IT COMES TOO SOLVING QUESTION....

harsh

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15 Apr 2012 19:54:57 IST

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ok , dude , i m nothing ,now be happy , u r very intelligent as compared to me , i agree with u ,now its ok

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