Mathematics is a very interesting subject. Although many people consider it to be the worst subject, it has many applications in the real world. For example, you could estimate the position of a cannonball relative to its starting position at a specific time (if you first receive the initial velocity). This kind of problem is a Vertical Motion Problem. For a vertical motion problem, you should know the different methods of solving a quadratic equation.
Our group picked a video of 2 people throwing a ball to each other (the link is below). https://drive.google.com/drive/folders/0ByYMD6W331LZaW5GejFxZ0VTMUk
After picking the video, we used an application called Tracker, and we traced the motion of the ball, not before we defined the radius of the ball to be 20 centimeters and placed the axis in an appropriate position.
After tracking the position of the ball, we place all of the values in the online graphing calculator Desmos. We made sure that we input the values in the table format, since a table would allow us to input the values we got from Tracker to Desmos.
After that, we put in the equation on top into Desmos as a function which would find the quadratic regression of the data values. A quadratic regression is kind of like a line of best fit, but instead of trying to find a linear function, it would have to find a quadratic equation. From that line of best fit, the graphic calculator would would find out the values of a, b and c in the quadratic equation. And what we found out was kind of interesting.
The quadratic equation that we used before this experiment for vertical motion problems was h(t) = -5t² + vt + h, where v is the initial velocity and h is the initial height above the ground. However, what we got instead for an equation is h(t) = -7.3216 t² + 52.686 t – 93.971.
For a start, the reason the constant of the quadratic equation is negative is because of the time we picked for measuring the motion of the ball in the air. Since we started tracking the motion of the ball not at the start of the video, the first value wouldn’t be on the y-axis. Therefore, the y-intercept of the quadratic regression would be negative, whereas it would be 0, if the part we measured was the first part of the video.
Next, the numerical value of b is just to compensate for the time we measured in the video. If we take out the constant in the quadratic equation, the vertex of the function would be higher than it was, and the graph would also move location in the graph. While experimenting with the graphing calculator, I manage to find out that the initial velocity of the ball was about 4 m/s. This makes the y-coordinates of the vertices between the 2 functions pretty close. But then I found out that you could also find out the initial velocity by dividing 52.686 with 3.237(3.237 being the x coordinate of the first value), before finding the square root of the quantity. The result was 4.034375001121689 m/s (round to 4.034). When I put that number into the function, the coordinates of the vertices were even closer.
Finally, the numerical value of a is because of also a measurement issue, but not related to time. For the graph, the first measurements we made were still when the ball was in the girl’s hands. This meant that the ball would be accelerating when she was pushing the ball. This would confuse the data then. But if we place the numerical value of a with -5 (did experimentation, round it to 5 for simplification purposes) instead of -7.3216, there wouldn’t be that many changes, since the changed initial velocity has the ability to really change the location of the the graph. The only difference is that the rate of increase of the slopes would be lower if we use -5.
So in the end, the ‘model’ equation from the data values is h(t) = -5t² + 4.034t from h(t) = -7.3216t² + 52.686t – 93.971.
To justify that they are similar, we would have to pick a time that we would use for each equation (remember that with the original equation, you would have to add 3.276, which is the x-intercept, to the time for it to work). After picking a time, you input the numerical value into t.
Justification: The time for justification would be 0.5 seconds
New equation: 0.77 meters
Old equation: 0.58 meters
So there is only a 0.19 meter difference between the 2 equations, and the new one had to be rounded for simplification purposes.
However, both equations work really well, it is just the old equation would be a lot more complicated to use: you would have to add a certain amount to the time you want for the vertical motion before you actually start doing the calculations. The new equation just tries to simplifies things so it is easier to calculate things. The y-coordinates of both of the vertices for the functions are very close, but for this one, the equations differ. I think this is because I didn’t round to a decimal place, which could haveover-simplified the answer. But the equations did match with the data values very well nonetheless, especially the old on (although it will be complicated to use).
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