Question

A person in an elevator accelerating upwards with an acceleration of 2 ms^-2, tosses a coin vertically upwards with a speed of 20 ms^-1. After how much time will the coin fall back into his hand? (g = 10 ms^-2)

Answer 1

As, v = u + at

t = \(\frac{20}{12}\) = \(\frac{5}{3}S\)

∴ Time of ascent = time of descent

∴ Total time after which coin falls back

t = \(\frac{2a}{g+a}\) = \(\frac{2\times 20}{10+2}\) = \(\frac{40}{12}\) = \(\frac{10}{3}\) = 3.33s.