To analyze the motion of the book, let's break down the problem into horizontal and vertical components, using the information provided.
1. Horizontal Components:
- Initial velocity ([tex]\(v_{0_x}\)[/tex]):
The initial velocity horizontally can be found using the cosine of the angle. The given speed is 12 m/s at an angle of [tex]\(35^\circ\)[/tex] below the horizontal.
[tex]\[
v_{0_x} = 12 \cos(35^\circ)
\][/tex]
From the given solution:
[tex]\[
v_{0_x} \approx 9.83 \, \text{m/s}
\][/tex]
- Acceleration ([tex]\(a_x\)[/tex]):
Since there is no horizontal acceleration:
[tex]\[
a_x = 0 \, \text{m/s}^2
\][/tex]
2. Vertical Components:
- Initial velocity ([tex]\(v_{0_y}\)[/tex]):
The initial velocity vertically can be found using the sine of the angle (note that it will be positive as we consider the downward direction as positive):
[tex]\[
v_{0_y} = 12 \sin(35^\circ)
\][/tex]
From the given solution:
[tex]\[
v_{0_y} \approx 6.88 \, \text{m/s}
\][/tex]
- Acceleration ([tex]\(a_y\)[/tex]):
The vertical acceleration is due to gravity:
[tex]\[
a_y = -9.8 \, \text{m/s}^2
\][/tex]
Now, summarizing the values in the table form as requested:
[tex]\[
\begin{tabular}{|l|c|c|}
\hline
& horizontal & vertical \\
\hline
Initial velocity & 9.83 \, \text{m/s} & 6.88 \, \text{m/s} \\
\hline
Acceleration & 0 \, \text{m/s}^2 & -9.8 \, \text{m/s}^2 \\
\hline
\end{tabular}
\][/tex]
Thus, the completed table is:
[tex]\[
\begin{tabular}{|l|c|c|}
\hline
& horizontal & vertical \\
\hline
Initial velocity & 9.83 & 6.88 \\
\hline
Acceleration & 0 & -9.8 \\
\hline
\end{tabular}
\][/tex]
