Problem 3: A rock of mass m = 13 kg is dropped from a height of h = 3.9 m into a pool of water. At a time of 1 = 1.96 s after hitting the
surface of the water, the rock's velocity has decreased by 50%.
What is the magnitude of the average force the rock experiences, in newtons, during the time t?
Numeric: A numeric value is expected and not an expression.

To solve this problem, we need to find the magnitude of the average force the rock experiences while it is in the water. Here’s the step-by-step process:

### Step 1: Determine the velocity of the rock just before it hits the water
The rock is dropped from a height \( h = 3.9 \, \text{m} \), and we can find its velocity just before impact using the equation of motion under free fall:

\[
v = \sqrt{2gh}
\]

Where:
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 3.9 \, \text{m} \)

\[
v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 3.9 \, \text{m}} = \sqrt{76.518} \approx 8.75 \, \text{m/s}
\]

So, the velocity of the rock just before it hits the water is approximately \( 8.75 \, \text{m/s} \).

### Step 2: Find the velocity of the rock after 1.96 seconds in water
The problem states that the velocity decreases by 50% after \( t = 1.96 \, \text{s} \). So, the velocity after 1.96 seconds is:

\[
v_{\text{after}} = 0.5 \times 8.75 \, \text{m/s} = 4.375 \, \text{m/s}
\]

### Step 3: Calculate the deceleration (negative acceleration)
The deceleration (or negative acceleration) can be found using the formula:

\[
a = \frac{\Delta v}{\Delta t}
\]

Where:
- \( \Delta v = v_{\text{after}} - v_{\text{initial}} = 4.375 \, \text{m/s} - 8.75 \, \text{m/s} = -4.375 \, \text{m/s} \)
- \( \Delta t = 1.96 \, \text{s} \)

\[
a = \frac{-4.375 \, \text{m/s}}{1.96 \, \text{s}} \approx -2.23 \, \text{m/s}^2
\]

### Step 4: Apply Newton's second law to find the average force
Newton's second law states that the force is the product of mass and acceleration:

\[
F = ma
\]

Where:
- \( m = 13 \, \text{kg} \)
- \( a \approx -2.23 \, \text{m/s}^2 \)

\[
F = 13 \, \text{kg} \times -2.23 \, \text{m/s}^2 \approx -28.99 \, \text{N}
\]

### Step 5: Magnitude of the average force
Since the problem asks for the magnitude, we take the absolute value:

\[
F_{\text{avg}} = | -28.99 \, \text{N} | \approx 29 \, \text{N}
\]

### Final Answer:
The magnitude of the average force the rock experiences during the time \( t \) is approximately **29 N**.

Would you like more details or have any questions about the solution?

Here are some related questions:
1. How would the force change if the mass of the rock was doubled?
2. What would the force be if the velocity reduction was only 25%?
3. How would the result change if the time interval was 3 seconds instead of 1.96 seconds?
4. What is the significance of deceleration in this context?
5. How would air resistance affect the velocity before hitting the water?

**Tip:** Always remember that when an object falls freely, its acceleration is due to gravity and independent of its mass.

A rock of mass m = 13 kg is dropped from a height of h = 3.9 m into a pool of water. At a time of t = 1.96 s after hitting the surface of the water, the rock's velocity has decreased by 50%. What is the magnitude of the average force the rock experiences, in newtons, during the time t?

Learn how to calculate the magnitude of the average force on a 13 kg rock dropped from 3.9 meters into water, with its velocity reduced by 50% over 1.96 seconds. This problem covers key physics concepts, including kinematics, forces, and Newton's Second Law, suitable for Grades 10-12.

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