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 of how long it will take for the coin to fall back into the person's hand in an accelerating elevator, we can follow these steps:

### Step 1: Understand the scenario
The elevator is accelerating upwards with an acceleration of \(2 \, \text{m/s}^2\). The person tosses the coin vertically upwards with an initial speed of \(20 \, \text{m/s}\). We need to find the time it takes for the coin to return to the person's hand.

### Step 2: Determine the effective acceleration
Since the elevator is accelerating upwards, we need to consider the effective acceleration acting on the coin. The acceleration due to gravity \(g\) is \(10 \, \text{m/s}^2\) (acting downwards), and the elevator's acceleration is \(2 \, \text{m/s}^2\) (acting upwards). Therefore, the net acceleration \(a\) acting on the coin is:
\[
...