To determine if the baseball travels over the fence, we need to find the height of the baseball when it reaches the horizontal distance where the fence is located and compare it to the height of the fence.

Step 1: Find the time when the baseball reaches the fence

The horizontal distance $x$ is given by the equation:

$x = 140t$

The fence is 320 feet away from home plate, so we set $x = 320$ and solve for $t$:

$140t = 320$

$t = \frac{320}{140} = \frac{32}{14} = \frac{16}{7} \approx 2.29 \text{ seconds}$

Step 2: Calculate the height of the baseball at $t = \frac{16}{7}$

The vertical position $y$ of the baseball is given by the equation:

$y = 3.1 + 40t - 16t^2$

Substitute $t = \frac{16}{7}$ into the equation:

$y = 3.1 + 40\left(\frac{16}{7}\right) - 16\left(\frac{16}{7}\right)^2$

First, calculate each term:

• $40\left(\frac{16}{7}\right) = \frac{640}{7} \approx 91.43$
• $\left(\frac{16}{7}\right)^2 = \frac{256}{49}$
• $16 \times \frac{256}{49} = \frac{4096}{49} \approx 83.59$

Now, plug these values into the equation for $y$:

$y = 3.1 + 91.43 - 83.59 \approx 10.94 \text{ feet}$

Step 3: Compare the height of the baseball with the height of the fence

The fence is 10 feet tall, and the height of the baseball when it reaches the fence is approximately 10.94 feet.

Conclusion

Since $10.94$ feet is greater than $10$ feet, the baseball does indeed travel over the fence.

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Would you like more details or have any questions? Here are 5 questions to consider:

1. What if the fence was 12 feet tall—would the ball still clear it?
2. How does the height equation change if the initial height were different?
3. Can you find the maximum height of the baseball during its flight?
4. What is the time at which the baseball reaches its maximum height?
5. How would the outcome change if the horizontal velocity was different?

Tip: When solving problems involving motion, always separate the horizontal and vertical components and use the correct equations to analyze them.