Question

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?

Answer 1

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

Step-by-step explanation:

Given that,

Speed = 42.0 m/s

Distance = 1.50 m

(a). We need to calculate the acceleration

Using equation of motion

`v^2=u^2+2as`

`a=(v^2-u^2)/(2s)`

Put the value in the equation

`a=(42.0^2-0)/(2*1.50)`

`a=588\ m/s^2`

(b). We need to calculate the time

Using equation of motion

`v = u+at`

`t=(v-u)/(a)`

`t=(42.0-0)/(588)`

`t=0.071\ sec`

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