The fastest measured pitched baseball left the pitcher’s hand at a speed of 45.0 m/s. If the pitcher was in contact with the ball over a distance of 1.50 m and produced constant acceleration, (a) what acceleration did he give the ball, and (b) how much time did it take him to pitch it?

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Juan Spiller · Accepted Answer

a) This is the analysis model particle under constant acceleration. vξ=0 In order to calculate the acceleration, we will use the following equation: vxf2=vξ2−2ax(xf−xi) 452=0+2ax(1.50) ax=675 ms2 b) We can calculate the time from the following equation: vxf=vxi+axt 45=0+675t t=0.067 s

The fastest measured pitched baseball left the pitcher’s hand at a spe

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