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        A lift is moving up with a constant acceleration=2.5 m/s^2 .when its upward velocity is 4 a boy in the lift tosses acoin imparting it an upward velocity =5 w.r.t. himself. His fingers at the moment of toss are midway b/w the floor and the ceiling , whose total height is 2.2m.After how much time will the coin hit the floor of the lift?Also find the distance travelled by the coin and its displacement in the earth frame
5 years ago

Arun Kumar
IIT Delhi
256 Points


Hello Student,

We’ll do this with in reference frame to lift

$\\a_{lift}=2.5m/s^2\vec j \\v_{lift}=4m/s\vec j \\v_{coin,rel}=5m/s\vec j \\dis_{rel}=-1.1mj \\a_{coin,rel}=-10-2.5=-12.5j \\so \\-1.1=5*t-12.5*t^2 =>t={5 \pm \sqrt{25+55} \over 25} \\take the positive t$

for earth frame use the vel relative to earth and the time we just found out.

Thanks & Regards
Arun Kumar
Btech, IIT Delhi


5 years ago
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• 731 Video Lectures
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions