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?
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?

a = 2 m/s2 is the rising acceleration of an elevator.

Gravitational acceleration, g = 10 m/s2

As a result, the net effective acceleration, a’ = (a + g) = 12 m/s2, is calculated.

In light of the coin’s effective velocity,

0 = v

t is the time it takes for the coin to reach its greatest height.

u = 20 m/s u = 20 m/s u = 20 m/

12 m/s2 = a’

As a result, v = u + at

After replacing the numbers in the preceding equation, t = 5/3 sec.

As a result, the coin’s total time to return after reaching maximum height is 3(1/3) sec.