Question

A person in an elevator accelerating upwards with an acceleration of $2 \mathrm {~ms} ^ { - 2 }$ , tosses coin vertically upwards with a speed of $20 \mathrm {~ms} ^ { - 1 }$ . After how much time will the coin fall back into his hand?

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

Answer 1

Answer: (C)

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

Answer 2

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

The coin will fall back into the person’s hand after t s $\therefore t = \frac { 2 v } { a + g } = \frac { 2 \times 20 m ^ { - 1 } } { ( 2 + 10 ) m s ^ { - 2 } } = \frac { 40 } { 12 } s = \frac { 10 } { 3 } \mathrm {~s}$