A person in an elevator accelerating upwards | vedclass

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Question

(C) The acceleration of the elevator is $a = 2\,m s^{-2}$ (upwards).The acceleration of the coin relative to the elevator is $a_{rel} = g + a = 10 + 2

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$\frac{10}{3}\,s$

Vedclass · Answer

(C) The acceleration of the elevator is $a = 2\,m s^{-2}$ (upwards).The acceleration of the coin relative to the elevator is $a_{rel} = g + a = 10 + 2 = 12\,m s^{-2}$ (downwards).The initial velocity of the coin relative to the elevator is $u = 20\,m s^{-1}$.The coin falls back into the hand when the relative displacement is $s = 0$.Using the equation of motion $s = ut + \frac{1}{2} a_{rel} t^2$:$0 = 20t - \frac{1}{2} (12) t^2$$6t^2 = 20t$$t = \frac{20}{6} = \frac{10}{3}\,s$.

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

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