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  • A person in an elevator accelerating upwards with an acceleration of 2 m/s^2, 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?

Answers (1)

Let ‘a’ be the upward acceleration of the elevator, a = 2m/s2

                g = 10 ms-2

Thus the net effective acceleration, viz., a’ is equal to (a+g)

Thus a’ = (2+10)

a’ = 12 ms-1

Now, considering the effective motion of the coin, v = 0 & t = time taken by the coin to achieve maximum height

u = 20 ms-1  and a’ = 12 ms-2

thus, v=u + at here a = a’

0 = 20 – 12t

Therefore, t = \frac{20}{12}s =\frac{5}{3} s

Thus, time of ascend = time of descend

Therefore, after achieving maximum height, total time to return in hand will be equal to

\frac{5}{3} + \frac{5}{3}

=\frac{10}{3}

= 3\frac{1}{3}  seconds.

Posted by

infoexpert24

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