A Fast Pitch. The fastest measured pitched baseball left the pitcher’s hand at a speed of $45.0m/s$. If the pitcher was in contact with the ball over a distance of$1.50m$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 fastest measured pitched basketball left the pitcher’s hand at $45.0m/s$.
• The pitcher was in contact with the ball over a distance of $1.50m$.

This law describes that a particular body will continue to remain in motion unless an external force acts on that body.

The time taken can be identified by dividing the distance and the average velocity of the baseball. Moreover, the acceleration of the ball can be identified by dividing the velocity by time.

(a)

From Newton’s law, the relationship between initial velocity, final velocity, and acceleration with distance traveledcan be expressed as:

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

Here, $u$is the initial velocity which is$0m/s$ , v is the final velocity, a is the acceleration of the ball, and s is the distance traveled by the ball.

Substituting the values in the above expression, we get-

$\begin{array}{rcl}{\left(45m/s\right)}^2& =& {\left(0m/s\right)}^2+2\times a\times 1.50m\\ a& =& \frac{2025m^2/s^2}{3m}\\ a& =& 675m/s^2\end{array}$

Thus, the acceleration of the ball is $675m/s^2$

Step 3: Determination of the time taken by the pitcher

(b)

From Newton’s first law, the distance traveled by the baseball is expressed as:

$v=u+at$

Here,$u$ is the initial velocity which is$0m/s$ , v is the final velocity, a is the acceleration of the ball, and t is the time taken.

Substituting the values in the above expression, we get-

$\begin{array}{rcl}& & \text{4.5m/s=0m/s+}\left(675m/s^2\right)\text{×t}\\ t& =& \frac{{\displaystyle \text{4.5m/s}}}{{\displaystyle \text{675}m/s}}\\ t& =& \text{0.0067s}\\ t& =& \text{6.7}\times {10}^{-3}s\end{array}$

Thus, the time taken by the pitcher to pitch the basketball is$t=\text{6.7}\times {10}^{-3}s$