Asked by josh
A major-league pitcher can throw a baseball in excess of 40.2 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches the catcher who is 17.7 m away from the point of release?
Answers
Answered by
drwls
Time to reach catcher = T = 17.7/40.2 = 0.440 seconds
Amoount of drop = (1/2)*g*T^2 = 0.95 meter
The ball can drop even more if a "sinker" is thrown. This involves aerodynamic forces that are not considered here.
Amoount of drop = (1/2)*g*T^2 = 0.95 meter
The ball can drop even more if a "sinker" is thrown. This involves aerodynamic forces that are not considered here.
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