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. $\frac{5}{3}$ $s$
• B. $\frac{3}{10}$ $s$
• C. $\frac{10}{3}$ $s$
• D. $\frac{3}{5}$ $s$

Answer 1

Answer: (C)

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.
$\therefore \quad$ $t=\frac{2 \,v}{a+g}$ $=\frac{2\times20 \,m \,s^{-1}}{\left(2+10\right)\,m \,s^{-2}}$ $=\frac{40}{12} \,s$ $=\frac{10}{3} \,s$