A person in an elevator accelerating upwards | vedclass

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(C) Let the frame of reference be the elevator. Since the elevator is accelerating upwards with $a = 2 \, m/s^2$,a pseudo force acts on the coin in th

Vedclass · Accepted Answer

$1.33$

Vedclass · Answer

(C) Let the frame of reference be the elevator. Since the elevator is accelerating upwards with $a = 2 \, m/s^2$,a pseudo force acts on the coin in the downward direction.The effective acceleration of the coin in the elevator's frame is $g_{eff} = g + a = 10 + 2 = 12 \, m/s^2$ (downwards).The initial velocity of the coin relative to the elevator is $u = 20 \, m/s$ (upwards).The coin falls back into the hand when its displacement relative to the elevator is $s = 0$.Using the equation of motion $s = ut + \frac{1}{2}at^2$:$0 = 20t - \frac{1}{2}(12)t^2$$0 = 20t - 6t^2$$6t^2 = 20t$$t = \frac{20}{6} = \frac{10}{3} \approx 3.33 \, s$.Wait,re-evaluating the options provided. If the question implies the coin is tossed relative to the ground,the time of flight is $T = \frac{2u}{g} = \frac{2 	imes 20}{10} = 4 \, s$. If the elevator is accelerating,the relative acceleration is $12 \, m/s^2$. The time taken is $t = \frac{2u}{g_{eff}} = \frac{2 	imes 20}{12} = \frac{40}{12} = 3.33 \, s$. Given the options,there might be a typo in the question's provided options or parameters. Assuming the standard calculation,$3.33 \, s$ is the correct result.

$A$ 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 (in $, s$)? $(g = 10 \, m/s^2)$

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