Question

A baseball pitcher throws a baseball with a speed of 43 m/s. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released. (Fig. 2-37).

Short answer

The average acceleration of the baseball is $264.1 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Step-by-step solution

Step 1. Relation between initial velocity, final velocity, acceleration, and displacement

The rate of change in the object’s displacement is known as the object’s velocity. It has both direction and magnitude.

The rate of change in the object’s distance is known as the object’s speed. It has magnitude but no direction.

Given data.

The initial velocity is $v_o = 0$as the baseball was started from rest.

The final velocity is $v_{} = 43 \frac{\mathrm{m}}{\mathrm{s}}$.

The displacement of the baseball is $\left(x - x_o\right) = 88 \mathrm{m}$.

Assumption.

Let the acceleration be $a$.

Now,you know that

role="math" localid="1642834448067" $v^2 = v_o^2 + 2a\left(x - x_o\right)\dots \left(i\right)$.

Step 2. Calculation of the acceleration

Now, substituting all the values in equation (i),

$\begin{array}{c}{v}^2 = v_o^2 + 2a\left(x - x_o\right)\\ {\left(43 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 = 0^2 + 2a \times 3.5 \mathrm{m}\\ \mathrm{a} = 264.1\frac{\mathrm{m}}{{\mathrm{s}}^2}\end{array}$

Therefore, the average acceleration of the baseball is $264.1\frac{\mathrm{m}}{{\mathrm{s}}^2}$.