A coin is thrown upwards with a velocity of 20 m/s relative to a person standing in a lift. If the lift is accelerating upwards with 2 m/s², calculate the time it takes for the coin to return to the p

• Acceleration of the lift upwards, $a_{lift} = 2 \text{ m s}^{-2}$.
• Initial velocity of the coin relative to the person, $u_{rel} = 20 \text{ m s}^{-1}$ (upwards).
• Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$ (downwards).

When an object is thrown upwards in a lift accelerating upwards, the effective acceleration experienced by the object relative to the lift is the sum of the gravitational acceleration and the lift's acceleration. This effective acceleration acts downwards.
Let's take the upward direction as positive.
The acceleration of the coin relative to the ground is $a_{coin, ground} = -g = -10 \text{ m s}^{-2}$.
The acceleration of the lift relative to the ground is $a_{lift, ground} = +a_{lift} = +2 \text{ m s}^{-2}$.
The acceleration of the coin relative to the lift, $a_{rel}$, is:

So, the effective acceleration acting on the coin relative to the lift is $12 \text{ m s}^{-2}$ downwards.

We want to find the time $t$ when the coin returns to the person's hand. This means the net displacement of the coin relative to the person is zero ($s_{rel} = 0$).
Using the kinematic equation:

Substitute the known values:

Factor out $t$ from the equation:

This gives two possible solutions:

1. $t = 0$ (This is the initial moment when the coin is tossed).
2. $20 - 6t = 0$ $6t = 20$ $t = \frac{20}{6} \text{ s}$ $t = \frac{10}{3} \text{ s}$

The time after which the coin returns to the person's hand is $\frac{10}{3} \text{ s}$.

Comparing this with the given options, it matches option (c).

The final answer is $\text{(c)}$.