To calculate the impulse imparted to the ball, we need to look at the change in momentum during the collision with the ground. The impulse $ I $ can be determined by the following relationship:

$ I = \Delta p $

Where:

• $ \Delta p $ is the change in momentum.
• Momentum $ p $ is defined as the product of the mass $ m $ of an object and its velocity $ v $.

Since the ball rebounds to the same height, its speed just before it hits the ground and just after it leaves the ground will be the same (ignoring air resistance), although the direction of the velocity will change.

Let's calculate the speed of the ball just before the impact. The ball drops from a height $ h $ with initial velocity $ u = 0 $ m/s. Using the kinematic equation for constant acceleration under gravity $ g $, we have:

$ v^2 = u^2 + 2gh $

Plugging in the values for $ u = 0 $, $ g = 10 $ m/s², and $ h = 10 $ m, we get:

$$ v^2 = 0^2 + 2 \cdot 10 \cdot 10 $$

$ v^2 = 200 $

$ v = \sqrt{200} $

$ v = 10\sqrt{2} $ m/s

Since the ball bounces back to the same height, its speed on the way up just after the collision will be $ 10\sqrt{2} $ m/s but in the opposite direction.

The change in velocity $ \Delta v $ is:

$$ \Delta v = v_{\text{final}} - v_{\text{initial}} $$

$$ \Delta v = 10\sqrt{2} - (-10\sqrt{2}) $$

$$ \Delta v = 20\sqrt{2} $$ m/s

Given that the mass $ m $ is 0.15 kg, we can now calculate $ \Delta p $:

$$ \Delta p = m \Delta v $$

$$ \Delta p = 0.15 \cdot 20\sqrt{2} $$

$ \Delta p = 3\sqrt{2} $

$$ \Delta p \approx 3 \cdot 1.414 $$

$$ \Delta p \approx 4.242 $$ kg m/s

The magnitude of the impulse imparted to the ball is therefore approximately 4.242 kg m/s, which most closely matches:

Option C: 4.2 kg m/s.