0

Guest

Courses New

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

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on General Physics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More