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Sagot :
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]
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]
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