Ncert Exercises & Solutions

A person in an elevator accelerating upwards with an acceleration of
2 ${ms}^{- 2}$ , tosses a coin vertically upwards with a speed of 20 m/s. After
how much time will the coin fall back into his hand? (g = 10 ${ms}^{- 2}$ )

Hint: Apply the concept of relative motion.

Step 1: Find

g_{eff}

Here, acceleration of the elevator (a) = 2 $m / s^{2}$ (upwards)
Acceleration due to gravity (g) = 10 $m / s^{2}$
Effective acceleration a' = g + a = 10+ 2 = 12 $m / s^{2}$ (here, acceleration is w.r.t. the lift)

Step 2: Find the time after which the coin falls back into the hand.

Initial speed of the coin (u) = 20 m/s

$\begin{array}{l} v = u + at \\ or 0 = 20 + (- 12) \times t \\ t = \frac{20}{12} = \frac{5}{3} s \end{array}$

Time of ascent = Tirne of desent

$∴$ Total time after which the coin falls back into the hand $= (\frac{5}{3} + \frac{5}{3}) s = \frac{10}{3} s = 3 .33 s$

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