To determine the missing height values [tex]\( A \)[/tex] and [tex]\( B \)[/tex] for times 5 and 6 seconds respectively, let’s analyze the motion of the baseball using the principles of projectile motion.
Here are the given data points for reference:
[tex]\[
\begin{array}{|c|c|}
\hline \text{Time (seconds)} & \text{Height (meters)} \\
\hline 0 & 0 \\
\hline 1 & 24.5 \\
\hline 2 & 39.2 \\
\hline 3 & 44.1 \\
\hline 4 & 39.2 \\
\hline 5 & A \\
\hline 6 & B \\
\hline
\end{array}
\][/tex]
To find the heights at specific times, we will use the projectile motion equation:
[tex]\[ h = h_0 + v_0 t - \frac{1}{2} g t^2 \][/tex]
where:
- [tex]\( h \)[/tex] is the height at time [tex]\( t \)[/tex]
- [tex]\( h_0 \)[/tex] is the initial height (0 meters in this case)
- [tex]\( v_0 \)[/tex] is the initial velocity (which we need to find)
- [tex]\( g \)[/tex] is the acceleration due to gravity (9.8 m/s[tex]\(^2\)[/tex])
Given points on the trajectory will help us determine the initial velocity. Let's use the points at [tex]\( t = 1 \)[/tex] second and [tex]\( t = 2 \)[/tex] seconds:
[tex]\[
h_1 = v_0 \cdot 1 - \frac{1}{2} \cdot 9.8 \cdot 1^2, \quad h_1 = 24.5
\][/tex]
[tex]\[
24.5 = v_0 - 4.9
\][/tex]
So,
[tex]\[
v_0 = 24.5 + 4.9 = 29.4 \, \text{m/s}
\][/tex]
We will use this initial velocity [tex]\( v_0 \)[/tex] to find the height at times [tex]\( A = 5 \)[/tex] seconds and [tex]\( B = 6 \)[/tex] seconds:
For [tex]\( t = 5 \)[/tex] seconds:
[tex]\[
A = 29.4 \cdot 5 - \frac{1}{2} \cdot 9.8 \cdot 5^2
\][/tex]
[tex]\[
A = 147 - 0.5 \cdot 9.8 \cdot 25
\][/tex]
[tex]\[
A = 147 - 122.5
\][/tex]
[tex]\[
A = 24.5 \, \text{meters}
\][/tex]
For [tex]\( t = 6 \)[/tex] seconds:
[tex]\[
B = 29.4 \cdot 6 - \frac{1}{2} \cdot 9.8 \cdot 6^2
\][/tex]
[tex]\[
B = 176.4 - 0.5 \cdot 9.8 \cdot 36
\][/tex]
[tex]\[
B = 176.4 - 176.4
\][/tex]
[tex]\[
B \approx 0 \, \text{meters}
\][/tex]
Therefore, the missing values in the table are:
[tex]\[
A = 24.5 \, \text{meters}
\][/tex]
[tex]\[
B = 0 \, \text{meters}
\][/tex]
