Question

Initially, a baseball was held by one pitcher.

The fastest measured pitched baseball left the pitcher’s hand at a speed of 45m/s. If the pitcher was in contact with the ball over a

distance of 1.5m and produced constant acceleration,



a)What acceleration did he give the ball?



b)How much time did it take him to pitch it?

Answer 1

Solve these:
v = at
s = 1/2 at^2

Answer 2

speed 0 to 45 m/s over 1.5 meters at constant a
how long was it in his hand?
v = a t, so v linear in time
so average speed = 45/2 =22.5 m/s
so time in hand= 1.5 /22.5 = 0.0667 seconds (ans part b)
v = a t still
45 = a (0.0667)
a = 675 m /s^2

by the way, 45m/s is 100 miles per hour. Really?