An elevator which is moving with an upward acceleration of "a m/s2”. A person drops a coin to measure the acceleration 'a m/s2'. Find 'a' if the coin took 0.5s to hit the ground and the coin was dropped from a height of 3 m.
A
10m/s2
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B
14m/s2
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C
5m/s2
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D
0m/s2
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Solution
The correct option is B14m/s2 Ulift=+U Ucoin=+U(same)
So ucoin/lift=U−U=0
So with respect to lift coin's initial velocity=0
Now alift=+a acoin=−g acoin/lift=−g−a
Now w.r.t. i.e., if I am the lift and I have eyes I will see the coin falling a fall of 3m. So the displacement will be -3 as in negative y-axis. Now, s=ut+12at2 −3=0+12(−g−a)(12)2(since,t=0.5sec)
⇒+3=(g+a)g
g+a=24
a=24-g=14 m/s2
So lift is going 14 m/s2 in upward direction.