Question

A person in an elevator accelerating upwards with an acceleration of 2m/sec^2, tosses a coin vertically upwards with a speed of 20m/sec. After how much time will a coin fall back into his hand?(g=10m/sec^2)
(1)1.3
(2)2.3
(3)3.3
(4)4.3
All are in seconds.

Answer 1

Answer on Question #48843 - Physics - Relativity

1. A person in an elevator accelerating upwards with an acceleration of $2\mathrm{m} / \mathrm{sec}^{\wedge}2$ , tosses a coin vertically upwards with a speed of $20\mathrm{m} / \mathrm{sec}$ . After how much time will a coin fall back into his hand? $(g = 10\mathrm{m} / \mathrm{sec}^{\wedge}2)$ : (1) 1.3; (2) 2.3; (3) 3.3; (4) 4.3. All are in seconds.

Let find the time of flying to the highest point (until stopping):

$$- g - a _ {0} = \frac {0 - v}{t}, \quad - g - a _ {0} = \frac {0 - v}{t _ {1}}, \quad t _ {1} = \frac {v}{g + a _ {0}}.$$

The time of returning is twice longer: $t = 2t_{1}$ , $\boxed{t = \frac{2v}{g + a_{0}}}$ .

Let check the dimension: $\left[t\right] = \frac{m}{s}:\frac{m}{s^2} = s$

Let evaluate the quantity: $t = \frac{2 \cdot 20}{10 + 2} = 3.33(s)$ .

Answer: 3.

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