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How an 85 mph pitcher

How an 85 mph pitcher. can throw 93 mph using math. Using ratios. Convert mph to ft/sec 1 mph= 5280 ft/sec 3600 Measure the distance the ball travels 60 ft 6 inches (60.5 ft) Determine how many seconds it takes the ball to reach home plate

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How an 85 mph pitcher

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  1. How an 85 mph pitcher can throw 93 mph using math

  2. Using ratios • Convert mph to ft/sec 1 mph= 5280 ft/sec 3600 • Measure the distance the ball travels 60 ft 6 inches (60.5 ft) • Determine how many seconds it takes the ball to reach home plate distance = seconds to home plate ft./sec

  3. Calculations • How many seconds does it take an 85mph pitch to get to home plate? • 85mph= (85)(5280/3600)ft/sec 124.67 ft/sec • Time to home plate=60.5/124.67 .485 seconds 60.5 feet

  4. The Flaws • Height of the mound, pitcher, and release point • Height of strike zone 6inches 74inches 24inches 10inches

  5. The Trigonometry of pitching • Use the Pythagorean theorem to solve for x c x ft. 5.5 ft. a b 60.5 ft. 2 2 2 a + b = c

  6. The Trigonometry of pitching • Use the Pythagorean theorem to solve for x 60.75 ft c x ft. 5.5 ft. a b 60.5 ft. 30.25 + 3660.25 = x 2 x =60.75

  7. Calculations Actual time to home plate 60.75/124.67 = .487 seconds Appears as an 84 mph fastball to the batter

  8. law of cosines x = 45 inches or 3.75 ft. 90(cos60) = x 56.9 ft. 7.5 ft. Pythagorean Theorem 4.5 ft. 90 inches 6.5 ft 56.75 ft 60 3.75 ft. x inches 6.5 – 2 = 4.5 ft.

  9. Calculations Actual time to home Plate 56.9/124.67 = .4564 Appears as an 90 mph fastball to the batter

  10. 55.1 ft. Law of Cosines 7.5cos(45) = x Pythagorean Theorem 5.5 ft. 5.5 ft = x 45 7.5 ft. 5.5 – 2 =3.5 ft. 3.5 ft. 45 x 5.5 ft. 60.5 – 5.5 =55 ft 55 ft.

  11. Calculations 60.5/.4419 = 136.91 ft/sec Actual time to home plate 55.1/124.67 = .4419 93.3 mph 136.91 * 3600 5280 = 93.3mph

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