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

Answer:

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

$v^2=u^2+2as$

$a=\dfrac{v^2-u^2}{2s}$

Put the value in the equation

$a=\dfrac{42.0^2-0}{2\times1.50}$

$a=588\ m/s^2$

(b). We need to calculate the time

Using equation of motion

$v = u+at$

$t=\dfrac{v-u}{a}$

$t=\dfrac{42.0-0}{588}$

$t=0.071\ sec$

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