Let's solve this step-by-step.

Problem Analysis

We have:

• A ball with mass $m = 25 \, \text{g} = 0.025 \, \text{kg}$,
• Initial height $h = 10 \, \text{m}$,
• The ball rebounds with $99\%$ of its incident speed after hitting the ground,
• The average net force during the collision with the ground is $F_{\text{avg}} = 45 \, \text{N}$,
• We need to find the time $\Delta t$ the ball was in contact with the ground.

Step 1: Determine the Incident Speed (Just Before Collision)

The ball falls freely from rest, so we can calculate the speed just before it hits the ground using energy conservation or kinematics. Here, I'll use energy conservation.

The potential energy at the height $h$ is fully converted to kinetic energy just before hitting the ground:

$mgh = \frac{1}{2}mv^2$

Solving for $v$:

$v = \sqrt{2gh}$

Using $g = 9.8 \, \text{m/s}^2$ and $h = 10 \, \text{m}$:

$v = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \, \text{m/s}$

So, the speed just before impact, $v_{\text{incident}}$, is $14 \, \text{m/s}$.

Step 2: Determine the Rebound Speed (After Collision)

The problem states the ball rebounds with $99\%$ of its incident speed. Therefore, the rebound speed $v_{\text{rebound}}$ is:

$v_{\text{rebound}} = 0.99 \times 14 = 13.86 \, \text{m/s}$

Step 3: Calculate the Change in Momentum (Impulse)

The change in momentum (impulse) is given by:

$\Delta p = m(v_{\text{rebound}} - (-v_{\text{incident}}))$

Here, $v_{\text{incident}}$ is downward (negative direction), and $v_{\text{rebound}}$ is upward (positive direction), so:

$\Delta p = m(v_{\text{rebound}} + v_{\text{incident}})$

Substitute the values:

$\Delta p = 0.025 \times (13.86 + 14) = 0.025 \times 27.86 = 0.6965 \, \text{kg} \cdot \text{m/s}$

Step 4: Calculate the Contact Time Using the Average Force

The impulse-momentum theorem states:

$\Delta p = F_{\text{avg}} \cdot \Delta t$

Solving for $\Delta t$:

$\Delta t = \frac{\Delta p}{F_{\text{avg}}}$

Substitute $\Delta p = 0.6965 \, \text{kg} \cdot \text{m/s}$ and $F_{\text{avg}} = 45 \, \text{N}$:

$\Delta t = \frac{0.6965}{45} = 0.015478 \, \text{s} \approx 0.0155 \, \text{s}$

Final Answer

The time the ball was in contact with the ground is approximately: $\Delta t \approx 0.0155 \, \text{s}$

Would you like more detailed explanations of any step or have further questions?

Related Questions

1. How would the contact time change if the rebound speed were 80% of the incident speed?
2. How does the mass of the ball affect the contact time with the ground?
3. What would be the incident speed if the ball fell from a different height, say 20 meters?
4. How would the answer change if the average net force was increased to 60 N?
5. What is the significance of the impulse-momentum theorem in collision problems?

Tip

Always check the units during calculations, especially when converting from grams to kilograms, to avoid mistakes in physics problems involving forces and momentum.