An elevator, in which a man is standing, is moving upward with a constant acceleration of 2m/s2 At some instant when speed of elevator is 10m/s, the man drops a coin from a height of 1.5m Find the time taken by the coin to reach the floor.
A
12sec
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1√2sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1√3sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1sec
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A12sec Step 1: Find the acceleration of coin with respect to elevator (→ace)
Given:
Acceleration of elevator (→aeg)=2m/s2
Acceleration of the coin →acg(=g)=10m/s2 →ace=→acg+→aeg →ace=(−10)−2 →ace=−12m/s2
Step 2:Find the velocity of coin with respect to elevator (→Vce)
As we know, Velocity of elevator=Velocity of coin=10m/s →Vce=→Vcg−→Veg=10−10 →Vce=0m/s
Step 3: Find the time taken by the coin to reach the floor.
As given, d=−1.5m
Initial velocity of coin u=0m/s
Therefore, from 2nd equation of motion drel=ut+12arelt2
Hence, drel=12arelt2
By putting the given values, we have, −1.5=12×(−12)×t2 1.5=12×12×t2 t=12sec
Final Answer: (c)