Physics, asked by Robotron, 1 year ago

a coin is released inside a lift at a height of 2 metre fom the floor of the lift. The height of the lift is 10 metre. The lift is moving with an acceleration of 11 m/s^2 downwards. The time after which coin will strike the lift is

Answers

Answered by tanya591
96

My first point is that you consider the question again because the acceleration of the lift is more than gravitational acceleration and moreover the unit is incorrect. If it is really having acceleration greater than g then the coin will hit the upper part of the lift.


a(rel)=accln. of lift - accln. of coin


a(rel)=11-10=1 m/s(2)


now distance of coin from upper roof= 8m


using s=ut+(1/2)at(square)


so u=0 as both are in the same velocity initially.


8=(1/2)*1*t(square)


t(square)=16


t=4.(ans)

Answered by adityameenaa
32

Brainly.in

What is your question?

Secondary SchoolPhysics 5+3 pts

A coin is released inside a lift at a height of 2 metre fom the floor of the lift. The height of the lift is 10 metre. The lift is moving with an acceleration of 11 m/s^2 downwards. The time after which coin will strike the lift is

Report by Robotron 31.07.2017

Answers

Adityameenaa · Helping Hand

Know the answer? Add it here!

tanya591

Tanya591 Expert

My first point is that you consider the question again because the acceleration of the lift is more than gravitational acceleration and moreover the unit is incorrect. If it is really having acceleration greater than g then the coin will hit the upper part of the lift.

a(rel)=accln. of lift - accln. of coin

a(rel)=11-10=1 m/s(2)

now distance of coin from upper roof= 8m

using s=ut+(1/2)at(square)

so u=0 as both are in the same velocity initially.

8=(1/2)*1*t(square)

t(square)=16

t=4.(ans)

Click to let others know, how helpful is it

4.0

8 votes

Report

Similar questions