On a baseball field, the pitcher’s mound is 60.5 feet from home plate. During practice, a batter hits a ball 216 feet. The path of the ball makes a 34° angle with the line connecting the pitcher and the catcher, to the right of the pitcher’s mound. An outfielder catches the ball and throws it to the pitcher. How far does the outfielder throw the ball?
207.4 ft 207.4 ft 224.3 ft 224.3 ft 169.3 ft 169.3 ft 198.7 ft
To solve this problem, we can first use trigonometry to find the horizontal distance the ball traveled from the pitcher's mound to the point where the outfielder caught it.
1. Find the horizontal distance the ball traveled before it was caught:
- cos(34°) = x / 216 ft
- x = 216 ft * cos(34°)
- x ≈ 179.23 ft
2. The outfielder then catches the ball and throws it to the pitcher. The distance between the outfielder and the pitcher's mound can be found using the Pythagorean theorem since the ball traveled in a straight line.
3. Let d be the distance the outfielder throws the ball, and z be the vertical distance the ball traveled:
- sin(34°) = z / 216 ft
- z = 216 ft * sin(34°)
- z ≈ 116.05 ft
4. Now, we can use the Pythagorean theorem to find the distance the outfielder throws the ball:
- d^2 = x^2 + z^2
- d^2 = (179.23 ft)^2 + (116.05 ft)^2
- d ≈ 207.4 ft
Therefore, the outfielder throws the ball approximately 207.4 feet.