Let's break down the problem step by step: --- ## Step 1: Was it a home run? **Given:** - The ball speed off the bat: \( v = 50 \) m/s - Launch angle: \( \theta = 30^\circ \) - Average outfield distance (to the fence): \( d = 100 \) m - Ignore air resistance. **Let's calculate how far the ball travels (range):** The horizontal range for projectile motion (no air resistance): \[ R = \frac{v^2 \sin 2\theta}{g} \] Where \( g = 9.8 \) m/s². Plug in the values: \[ R = \frac{(50)^2 \cdot \sin(60^\circ)}{9.8} \] \[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx .866 \] \[ R = \frac{250 \times .866}{9.8} \approx \frac{2165}{9.8} \approx 221 \text{ m} \] **Explanation:** The ball would travel 221 m, but the outfield fence is only 100 m away. So, **it was definitely a home run**. --- ## Step 2: Could the outfielder get in position to make the catch? **More info:** - Fielder can run at a maximum speed of 10 m/s. - Ball travels 221 m, but the fence is at 100 m. Let's find how long it takes the ball to reach 100 m horizontally. Horizontal velocity: \[ v_x = v \cos \theta = 50 \cos 30^\circ = 50 \times .866 = 43.3 \text{ m/s} \] Time to reach 100m: \[ t = \frac{d}{v_x} = \frac{100}{43.3} \approx 2.31 \text{ s} \] **How far can the fielder run in that time?** \[ \text{Distance} = \text{speed} \times \text{time} = 10 \times 2.31 = 23.1 \text{ m} \] **Explanation:** The fielder can only cover 23.1 m in the time it takes the ball to reach the fence, but the ball is already traveling much farther and faster. Unless the fielder was already standing right at the fence, they would **not** be able to make the catch. --- ## **Final Answer** 1. **Yes, it was a home run** (the ball travels 221 m, much farther than the 100 m fence). 2. **No, the outfielder could not get in position to make the catch** (they can only run 23.1 m in the time it takes the ball to reach the fence).