A person in an elevator accelerating upwards with an acceleration of 2 m/s2, tosses a coin vertically upwards with a speed of 20 m/s. After how much time will the coin fall back into his hand?
The upward acceleration of an elevator, a = 2 m/s2
Acceleration due to gravity, g = 10 m/s2
Therefore, the net effective acceleration, a’ = (a + g) = 12 m/s2
Considering the effective motion of the coin,
v = 0
t = time taken by the coin to achieve maximum height
u = 20 m/s
a’ = 12 m/s2
Therefore, v = u + at
Where, t = 5/3 sec after substituting the values in the above equation.
Therefore, the total time taken by the coin to return after achieving maximum height is 3(1/3) sec.
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