JEE Physics 2023 Question with Solution

Given: Initial height is $h = 20 \, \text{m}$ and coefficient of restitution is $e = 0.5$.

Find: The rebound height $h'$ after collision with the floor.

For a ball rebounding vertically, the rebound height is given by

$$h' = e^2 h$$

Substituting the values,

$$h' = (0.5)^2 \times 20$$ $$h' = 0.25 \times 20 = 5 \, \text{m}$$

Therefore, the ball rebounds to a height of $5 \, \text{m}$.

Given: The ball is dropped from $20 \, \text{m}$ and the coefficient of restitution is $0.5$.

Find: The height reached after the bounce.

The hint states that rebound height depends on the square of the coefficient of restitution. Hence,

$$\frac{h'}{h} = e^2$$

So,

$$h' = e^2 h$$

Putting $e = 0.5$ and $h = 20 \, \text{m}$,

$$h' = (0.5)^2 \times 20$$ $$h' = 0.25 \times 20$$ $$h' = 5 \, \text{m}$$

Therefore, the required numerical value is 5.