A person in an elevator accelerating upwards with an acceleration of 2

1.	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» 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.

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»

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.