To determine the new position of the nerf basketball at $t = 0.12$ s and $t = 0.14$ s, we'll use the principles of kinematics and Newton's second law. Here's the plan:

1. Calculate the acceleration at $t = 0.10$ s and $t = 0.12$ s using the net forces provided.
2. Find the velocity of the ball at $t = 0.12$ s and $t = 0.14$ s.
3. Determine the position of the ball at $t = 0.14$ s using the velocity at $t = 0.12$ s and the acceleration.

Step 1: Calculate Acceleration

Using Newton's second law:

$\vec{a} = \frac{\vec{F}_{\text{net}}}{m}$

For $t = 0.10$ s:

• $\vec{F}_{\text{net}} = (-0.0184, -0.0224, 0)$ N
• $m = 0.002$ kg

For $t = 0.12$ s:

• $\vec{F}_{\text{net}} = (-0.0138, -0.0197, 0)$ N

Step 2: Update Velocity

The velocity at $t = 0.12$ s can be calculated using:

$\vec{v}_{\text{new}} = \vec{v}_{\text{old}} + \vec{a} \Delta t$

Similarly, for $t = 0.14$ s, use the velocity from $t = 0.12$ s.

Step 3: Update Position

The new position can be found using:

$\vec{r}_{\text{new}} = \vec{r}_{\text{old}} + \vec{v} \Delta t$

We'll compute these values step by step.The position of the nerf basketball at $t = 0.14$ s will be approximately $(0.390, 0.201, 0.0)$ meters.

Would you like further details on the calculations, or do you have any additional questions?

Here are five related questions you might consider:

1. How would the position change if the air resistance was different?
2. Can you explain how the force of gravity is integrated into these calculations?
3. What would happen if the ball was thrown vertically instead?
4. How does air resistance affect the motion of the ball over time?
5. What is the significance of using small time intervals in such calculations?

Tip: When dealing with forces and motion, breaking the problem into small time steps can significantly improve the accuracy of the results.