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Grade 11General Physics

an elevator car whose floor to ceiling distance is equal to 2.7m starts from rest with a constant acceleration of 1.2 ms-2 . two seconds later , a bolt drops from car's ceiling.the time taken by bolt to hit floor of elevtor is..plz say..

Profile image of shyam sundar
14 Years agoGrade 11
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2 Answers

Profile image of abhishek upadhyay
ApprovedApproved Tutor Answer14 Years ago

mention whether elevator is going up or down

this problem will be easier if u solve it in frame of elevator

when bolt gets detached its initial  velocity w.r.t. elevator is 0

abE = ab(g) - aE(g)

       =g - a               if elevator is going down taking down +ve

      or    g+a                "   "      "      "        "  up     "        "      "

 now   2.7 = ubE t   +   abEt2/2

             = 0  +   abEt2/2 

   imp - block moves 2.7 w.r.t elevator

Profile image of Atharva Baste
8 Years ago
V0=0
T=?
S=2.7m
A=g+a(lift)=9.8+1.2=11m/s
Using,
s=Vo(t)+(0.5)(A)(T)(T)
=> 2.7/(.5) = 0 (t) + (11)(T)(T)
=>T=.70 sec