A baseball leaves a pitcher’s hand horizontally at a speed of$\mathbf{161}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}$. The distance to the batter islocalid="1654591573193" ${}^{\mathbf{18}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{m}}$. (a)How long does the ball take to travel the first half of that distance? (b) The second half? (c)How far does the ball fall freely during the first half? (d) During the second half? (e)Why aren’t the quantities in (c) and (d) equal?

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

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

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,

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

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.