To solve this problem, we need to calculate the impulse imparted to the ball.$\\$Impulse is defined as the change in momentum of an object.$\\$Given:$\\$• Mass of the ball $m = 0.15$ kg$\\$• Height from which the ball is dropped $h = 10$ m$\\$• Acceleration due to gravity $g = 10$ m/s$^2 \\$First, calculate the velocity of the ball just before it strikes the ground.$\\$Using the kinematic equation:$\\ v^2 = u^2 + 2gh \\$Since the ball is dropped, the initial velocity $u = 0. \\ v^2 = 0 + 2 \cdot 10 \cdot 10 \\ v^2 = 200 \\ v = \sqrt{200} \\ v = 10\sqrt{2}$ m/s$\\$The velocity just before striking the ground is $10\sqrt{2}$ m/s downward.$\\$Since the ball rebounds to the same height, the velocity just after rebounding is also $10\sqrt{2}$ m/s upward.$\\$The change in velocity $\Delta v$ is:$\\ \Delta v = v_{final} - v_{initial} \\ \Delta v = 10\sqrt{2} - (-10\sqrt{2}) \\ \Delta v = 20\sqrt{2}$ m/s$\\$The change in momentum $\Delta p$ is given by:$\\ \Delta p = m \cdot \Delta v \\ \Delta p = 0.15 \cdot 20\sqrt{2} \\ \Delta p = 3\sqrt{2}$ kg m/s$\\$Approximating $\sqrt{2} \approx 1.414,$ we get:$\\ \Delta p \approx 3 \cdot 1.414 \\ \Delta p \approx 4.242$ kg m/s$\\$Therefore, the magnitude of the impulse imparted to the ball is approximately $4.2$ kg m/s.$\\$This corresponds to Option 2.