91Ó°ÊÓ

It is given that, the horizontal component of the initial velocity of the baseball is,

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{x}}=161\mathrm{km}/\mathrm{h} \\ =161×\frac{5}{18} \\ =44.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Distance to the batter is${}^{\mathrm{x}=18.3\mathrm{m}}$

This problem is based on kinematic equations that describe the motion of an object with constant acceleration.

Using these equations, the vertical displacement and time can be found.

Formula:

According to the Newton’s second kinematic equation, the total displacement in y direction is given by,

$∆y=v_{0y}\left(\begin{array}{c}t\end{array}\right)+\frac{1}{2}a{t}^2$ (i)

Consider here displacements during the first and second half will be${}^{y_1}$and ${}^{y_2}$

In the first half, the displacement of the ball is

$$\begin{gathered} ∆\mathrm{x}=\frac{18.3}{2} \\ =9.15\mathrm{m} \end{gathered}$$

The horizontal displacement of the baseball in the first half is,

$∆\mathrm{x}={\mathrm{v}}_{0\mathrm{x}}×\mathrm{t}$

So, the time taken by the baseball to travel the first half of the distance,

$$\begin{gathered} \mathrm{t}=\frac{∆\mathrm{x}}{{\mathrm{v}}_{0\mathrm{x}}} \\ =\frac{9.15}{44.7} \\ \mathrm{t}=0.205\mathrm{s} \end{gathered}$$

For the second half, the velocity and the displacement is the same. So,the time taken by the baseball to travel the second half is same.

$t=0.205s$

Using equation (i), the vertical displacement of the ball during the first half can be calculated as,

$$\begin{gathered} {\mathrm{y}}_1=0+\frac{1}{2}\left(\begin{array}{c}-9.8\end{array}\right){\left(\begin{array}{c}0.205\end{array}\right)}^2 \\ \left|{\mathrm{y}}_1\right|=0.205\mathrm{m} \end{gathered}$$

The total vertical displacement of the baseball is,

$$\begin{gathered} ∆\mathrm{y}=0+\frac{1}{2}\left(\begin{array}{c}-9.8\end{array}\right){\left(\begin{array}{c}0.409\end{array}\right)}^2 \\ \left|∆\mathrm{y}\right|=0.820\mathrm{m} \end{gathered}$$

Therefore, the vertical displacement of the ball during the second half is,

$$\begin{gathered} {\mathrm{y}}_2=0.820-0.205 \\ {\mathrm{y}}_2=0.615\mathrm{m} \end{gathered}$$

The displacement of the baseball during the first half and the second half is different because the proportionality relation between vertical displacement and t is not linear. Also, the initial velocity for first half is not equal to the initial velocity for the second half.