A spherical body of mass $$100 \text{ g}$$ is dropped from a height of $$10 \text{ m}$$ from the ground. After hitting the ground, the body rebounds to a height of $$5 \text{ m}$$. The impulse of force imparted by the ground to the body is given by: (given $$g = 9.8 \text{ m s}^{-2}$$)

A spherical body of mass $$m = 100 \text{ g} = 0.1 \text{ kg}$$ is dropped from height $$10 \text{ m}$$ and rebounds to height $$5 \text{ m}$$.

Find the velocity just before hitting the ground: using $$v = \sqrt{2gh}$$:

$$v_1 = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \text{ m/s}$$ (downward)

Find the velocity just after rebounding: $$v_2 = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 7\sqrt{2} \text{ m/s}$$ (upward)

Calculate the impulse: impulse equals the change in momentum. Taking upward as positive:

$$J = m(v_2 - (-v_1)) = m(v_2 + v_1)$$

$$J = 0.1 \times (7\sqrt{2} + 14) = 0.1 \times (9.899 + 14) = 0.1 \times 23.899$$

$$J \approx 2.39 \text{ kg m s}^{-1}$$

The correct answer is Option (4): $$2.39 \text{ kg m s}^{-1}$$.