To determine how many rows $N$ are needed in the table to model the ball's motion within a 2% error, we will first compare the position calculations using two different approaches: an approximate method that neglects the term $\frac{1}{2}a(\Delta t)^2$, and the exact position formula.

Step 1: Position formula derivation

The exact position $x(t)$ at any time $t$ under constant acceleration can be found using the equation of motion:

$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$

Since we know the initial conditions:

• $x_0 = 6.7 \, \text{m}$,
• $v_0 = 0 \, \text{m/s}$,
• $a = 2.6 \, \text{m/s}^2$,

the position formula simplifies to:

$x(t) = 6.7 + \frac{1}{2} \cdot 2.6 \cdot t^2 = 6.7 + 1.3 t^2$

This is the exact position formula.

Step 2: Approximate formula

In the approximate method, we neglect the term $\frac{1}{2} a (\Delta t)^2$ that appears in the exact position expression. This gives an approximate position at time $t$ based only on the initial position and velocity (which is zero in this case):

$x_{\text{approx}}(t) = x_0 = 6.7 \, \text{m}$

This means that if we divide the time into intervals, we approximate the position based on the initial position, without accounting for acceleration properly.

Step 3: Error due to the approximation

The error at any time $t$ is the difference between the exact and approximate positions:

$\text{Error}(t) = x(t) - x_{\text{approx}}(t) = \left(6.7 + 1.3 t^2\right) - 6.7 = 1.3 t^2$

We need this error to be within 2% of the total displacement over the time period $t = 3.8 \, \text{s}$. To calculate this:

1. Find the exact position at $t = 3.8 \, \text{s}$: $x(3.8) = 6.7 + 1.3 \cdot (3.8)^2 = 6.7 + 1.3 \cdot 14.44 = 6.7 + 18.772 = 25.472 \, \text{m}$

2. The acceptable error is 2% of the total displacement: $2\% \text{ of displacement} = 0.02 \cdot (25.472 - 6.7) = 0.02 \cdot 18.772 = 0.37544 \, \text{m}$

So, we want the error due to neglecting $\frac{1}{2} a (\Delta t)^2$ to be less than or equal to 0.37544 m over any time interval $\Delta t$.

Step 4: Find time interval $\Delta t$

The error per time step is given by $\frac{1}{2} a (\Delta t)^2$:

$\text{Error per step} = \frac{1}{2} \cdot 2.6 \cdot (\Delta t)^2 = 1.3 (\Delta t)^2$

We set this error equal to the acceptable error per time step (0.37544 m):

$1.3 (\Delta t)^2 = 0.37544$

Solving for $\Delta t$:

$(\Delta t)^2 = \frac{0.37544}{1.3} \approx 0.28803$ $\Delta t \approx \sqrt{0.28803} \approx 0.5366 \, \text{s}$

Step 5: Calculate the number of rows $N$

The total time interval is 3.8 seconds. If each time step $\Delta t \approx 0.5366 \, \text{s}$, the number of rows $N$ is:

$N = \frac{3.8}{0.5366} \approx 7.08$

Since the number of rows must be an integer, we round up to the nearest whole number:

$N = 8$

Final Answer:

You will need 8 rows in the table to model the motion within the required 2% error.

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Would you like more details or clarifications on any part of the solution? Here are five related questions to explore further:

1. How would the result change if the acceptable error was reduced to 1%?
2. How is the concept of significant figures applied in this problem?
3. What role does acceleration play in determining the total displacement?
4. How would changing the acceleration affect the number of rows needed in the table?
5. How can this method be extended to two-dimensional motion?

Tip: Always check if simplifying assumptions, like neglecting terms, still meet accuracy requirements in motion problems.