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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..

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..

Grade:11

2 Answers

abhishek upadhyay
23 Points
11 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

Atharva Baste
37 Points
5 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

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