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?

Question

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? (Take g = 10 $m s^{- 2}$ )

Smart Achievers · Accepted Answer

(3)

Here, $v = 20 {m s}^{- 1}, a = 2 {m s}^{- 2}, g = 10 {m s}^{- 2}$

The coin will fall back into the person's hand after $t$ s.

$∴$ $t = \frac{2 v}{a + g} = \frac{2 \times 20 {m s}^{- 1}}{(2 + 10) {m s}^{- 2}} = \frac{40}{12} s = \frac{10}{3} s$

Solution

Q.
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 (Take g = 10 $m s^{- 2}$ )

1 $\frac{5}{3} s$

2 $\frac{3}{10} s$

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

4 $\frac{3}{5} s$

Ans.
(3)

Here, $v = 20 {m s}^{- 1}, a = 2 {m s}^{- 2}, g = 10 {m s}^{- 2}$

The coin will fall back into the person's hand after $t$ s.

$∴$ $t = \frac{2 v}{a + g} = \frac{2 \times 20 {m s}^{- 1}}{(2 + 10) {m s}^{- 2}} = \frac{40}{12} s = \frac{10}{3} s$

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Solution

Q. 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 (Take g = 10 m s-2)

1 53 s

2 310 s

3 103 s

4 35 s

Ans. (3) Here, v=20 m s-1, a=2 m s-2, g=10 m s-2 The coin will fall back into the person's hand after t s. ∴ t=2va+g=2×20 m s-1(2+10) m s-2=4012 s=103 s