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

Explanation

To solve the problem, we need to analyze the motion of the coin tossed in the accelerating elevator. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the scenario
A person in an elevator is accelerating upwards with an acceleration of \( a = 2 \, \text{m/s}^2 \). The coin is tossed upwards with an initial velocity \( u = 20 \, \text{m/s} \). We need to find the time \( t \) it takes for the coin to return to the person's hand.

### Step 2: Determine the effective acceleration
Since the elevator is accelerating upwards, the effective acceleration acting on the coin (when viewed from the elevator's frame of reference) will be the sum of gravitational acceleration \( g \) and the elevator's acceleration \( a \).

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