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Let's solve the problem step by step.

**Problem Summary:**
1. A 25 g bouncy ball is dropped from a height of 10 m.
2. It rebounds with 99% of its incident speed.
3. We need to find the time the ball was in contact with the ground.
4. The average net force during the collision with the ground is 45 N.

### Step 1: Convert Mass to kg
The mass of the ball is given as 25 g. Convert this to kilograms:
\[
m = 25 \, \text{g} = 0.025 \, \text{kg}
\]

### Step 2: Calculate the Speed Just Before Impact
Using the equation of motion for a free-falling object starting from rest:
\[
v = \sqrt{2gh}
\]
where:
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 10 \, \text{m} \).

Substitute the values:
\[
v = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} \approx 14 \, \text{m/s}
\]

So, the speed of the ball just before impact is approximately \( 14 \, \text{m/s} \).

### Step 3: Calculate the Rebound Speed
The ball rebounds with 99% of its incident speed:
\[
v_{\text{rebound}} = 0.99 \times 14 \approx 13.86 \, \text{m/s}
\]

### Step 4: Calculate the Change in Momentum (Impulse)
The change in momentum (impulse) during the collision is given by:
\[
\Delta p = m \times (v_{\text{rebound}} - (-v))
\]
where:
- \( v_{\text{rebound}} = 13.86 \, \text{m/s} \) (rebound speed),
- \( v = 14 \, \text{m/s} \) (initial speed just before impact).

Substitute the values:
\[
\Delta p = 0.025 \times (13.86 + 14) = 0.025 \times 27.86 = 0.6965 \, \text{kg m/s}
\]

### Step 5: Calculate the Contact Time
The average net force (\( F \)) during the collision is given as 45 N. Using the impulse-momentum theorem:
\[
\Delta p = F \cdot \Delta t
\]
where \( \Delta t \) is the contact time. Rearrange to solve for \( \Delta t \):
\[
\Delta t = \frac{\Delta p}{F} = \frac{0.6965}{45} \approx 0.0155 \, \text{s}
\]

### Final Answer:
The ball was in contact with the ground for approximately **0.0155 seconds**.

---

Would you like further explanation on any of the steps?

Here are some related questions that might interest you:
1. How would the contact time change if the rebound speed were different?
2. What would be the effect of a different height on the impact speed?
3. How does increasing the mass of the ball affect the impulse?
4. Can we determine the coefficient of restitution from this problem?
5. What role does air resistance play in real-life scenarios like this?

**Tip:** The impulse-momentum theorem is essential for understanding collisions, as it relates force, time, and change in momentum in a straightforward manner.

Starting from rest, a 25 g bouncy ball falls from a height of 10 m onto a flat, horizontal surface and rebounds with 99% of its incident speed. Find the amount of time that the ball was in contact with the ground if the average net force on the ball during the collision is 45 N.

Learn how to calculate the contact time for a 25 g ball falling from 10 m, rebounding with 99% of its speed. This problem covers key concepts in kinematics and impulse-momentum. Ideal for Grades 10-12 physics students.

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