The fastest measured pitched baseball left the pitcher's hand at a speed of 42.0 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?

The acceleration and time are 588 m/s² and 0.071 sec respectively.

Explanation:

Given that,

Speed = 42.0 m/s

Distance = 1.50 m

(a). We need to calculate the acceleration

Using equation of motion

Put the value in the equation

(b). We need to calculate the time

Using equation of motion

Hence, The acceleration and time are 588 m/s² and 0.071 sec respectively.