1. Given data

• Mass of sphere: $m=100\text{ g}=0.1\text{ kg}$
• Dropped from height: $h_1=10\text{ m}$
• Rebounds to height: $h_2=5\text{ m}$
• Acceleration due to gravity: $g=9.8\text{ m s}^{-2}$

We need the impulse imparted by the ground.

Impulse equals change in momentum during collision: $J=\Delta p=m(v_{\text{after}}-v_{\text{before}})$

Since the direction reverses, we must use signs carefully.

2. Velocity just before hitting the ground

Using $v^2=u^2+2gh$ with $u=0$, $v_1=\sqrt{2gh_1}=\sqrt{2\cdot 9.8\cdot 10}=\sqrt{196}=14\text{ m/s}$

This velocity is downward, so taking upward as positive: $v_{\text{before}}=-14\text{ m/s}$

3. Velocity just after rebounding from the ground

To rise to height $5\text{ m}$, $v_2=\sqrt{2gh_2}=\sqrt{2\cdot 9.8\cdot 5}=\sqrt{98}\approx 9.9\text{ m/s}$

This is upward, so $v_{\text{after}}=+9.9\text{ m/s}$

4. Calculate impulse

$J=m(v_{\text{after}}-v_{\text{before}})$ $J=0.1\,[9.9-(-14)]$ $J=0.1\times 23.9$ $J=2.39\text{ kg m s}^{-1}$

5. Match with options

$2.39\text{ kg m s}^{-1}$ corresponds to Option B.

6. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.