A major-league pitcher can throw a ball in excess of 39.1 m/s. If a ball is thrown horizontally at this spe?
A major-league pitcher can throw a ball in excess of 39.1 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?
- GeronimoLv 79 years agoFavorite Answer
Dx = horizontal distance traveled = 17.0 meters
Vx = horizontal speed = 39.1 m/sec = 87.46 mi/h
Time of Flight = t = (Dx) ⁄ (Vx) = (17.0) ⁄ (39.1) = 0.43478 sec
h = height relative to initial height
Vyi = initial vertical speed = 0 ... (ball thrown horizontally)
h = (½) • g • t² + (Vyi) • t
h = (½) • (-9.81) • (0.43478)²
h = - 0.927 meters ... almost a meter drop-off
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