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)`)

Options

A. `(5)/(3)s`
B. `(3)/(10)s`
C. `(10)/(3)s`
D. `(3)/(5)s`

Explanation

To solve the problem of when the coin will fall back into the person's hand in an upward-accelerating elevator, we can follow these steps:

### Step 1: Identify the variables
- Initial velocity of the coin, \( u = 20 \, \text{m/s} \)
- Acceleration of the elevator, \( a = 2 \, \text{m/s}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Determine the effective acceleration
Since the elevator is accelerating upwards, the effective acceleration acting on the coin (which is thrown upwards) will be the acceleration due to gravity minus the acceleration of the elevator. Thus, we have:

\[ a' = -g + a = -10 + 2 = -12 \, \text{m/s}^2 \]

### Step 3: Use the kinematic equation
We will use the kinematic equation to find the time taken for the coin to reach its highest point (where its velocity becomes zero):

\[ v = u + a' t \]

At the highest point, the final velocity \( v = 0 \). Substituting the values:

\[ 0 = 20 + (-12)t \]

### Step 4: Solve for time \( t \)
Rearranging the equation gives:

\[ 12t = 20 \implies t = \frac{20}{12} = \frac{5}{3} \, \text{s} \]

This time \( t \) is the time taken to reach the highest point.

### Step 5: Calculate the total time
Since the time taken to ascend is equal to the time taken to descend, the total time \( T \) for the coin to fall back into the person's hand is:

\[ T = t + t = 2t = 2 \times \frac{5}{3} = \frac{10}{3} \, \text{s} \]

### Step 6: Final answer
Thus, the total time taken for the coin to fall back into the person's hand is:

\[ T = \frac{10}{3} \, \text{s} \approx 3.33 \, \text{s} \]