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  • A lift starts moving down with a constant accelaration <img alt="15m/s^2" src="http://entrancecorner.oncodecog

A lift starts moving down with a constant accelaration 15m/s^2. A coin initially is at rest on the floor of the lift. The height of the lift is 2.5m. How much time (in seconds) will the coin take to touch the ceiling of the lift? (g=10m/s^2)

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Given-

Height of lift h=2.5m, downward acceleration of the lift, a_{l}=15m/s^2

Since the acceleration of the lift is greater than the acceleration due to gravity and in the downward direction, the coin will lose contact with the floor of the lift immediately after the motion begins. The motion of the coin will be free fall under gravity.

Therefore, the acceleration of the coin- a_{c}=g=10m/s^2

Acceleration of the lift with respect to coin-

a_{lc}=a_{l}-a_{c}=15-10=5m/s^2

Displacement of lift with respect to the coin- s_{lc}=2.5m

The initial velocity of lift with respect to coin, u_{lc}=0

Applying the second equation of motion for uniformly accelerated motion-

\\ s_{lc}=u_{lc}t+\frac{1}{2}a_{lc}t^2\\ \Rightarrow 2.5=0+\frac{1}{2}\times5\times t^2\\ \Rightarrow t=\sqrt{\frac{2 \times 2.5}{5}} =1s

Posted by

jitender.kumar

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