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)

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) (A) (5/3) s (B) (3/10) s (C) (10/3) s (D) (3/5) s

Tardigrade Admin · Accepted Answer

Here, v=20 m s-1, a=2 m s-2, g=10 m s-2 The coin will fall back into the personâs hand after t s. ∴ t=(2 v/a+g) =(2×20 m s-1/(2+10) m s-2) =(40/12) s =(10/3) s

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

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

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

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

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

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

Q. A person in an elevator accelerating upwards with an acceleration of 2m s−2, tosses a coin vertically upwards with a speed of 20m s−1. After how much time will the coin fall back into his hand? (Takeg=10ms−2)

35​ s

103​ s

310​ s

53​ s