What are the equations of these free fall/drag graphs?

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The discussion centers on analyzing the velocity and deceleration graphs of a basketball in free fall, launched at a 45-degree angle with an initial velocity of 10 m/s. The velocity graph shows the basketball's speed asymptotically approaching a terminal velocity of -20 m/s due to air resistance, while the second graph represents the deceleration caused by this drag. Participants clarify that the deceleration should be referred to as acceleration and discuss the plateau in the acceleration graph, attributing it to decreasing drag as the basketball reaches the peak of its trajectory. The relevant equation for drag force, Fd = 0.5 * Cd * rho * A * v^2, is acknowledged as the basis for the graphs, with a focus on determining its applicability to the observed data. The conversation emphasizes the importance of precise terminology and understanding the underlying physics to interpret the graphs accurately.
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Homework Statement


The screenshots of the graphs are attached. The first graph is a velocity graph of a basketball with respect to time, and it is assumed that it has been launched at an angle of 45 degrees and an initial velocity of 10 m/s. Due to air resistance, which I was able to include using Euler method, spreadsheet, and LoggerPro, the velocity asymptotically approaches -20m/s, which actually is the terminal velocity of a falling basketball.

However, I don't know what kind of equation is behind that graph.

The second graph is the deceleration from the air resistance, and I don't know what the equation for that one is either.
Screen Shot 2018-09-30 at 12.27.06 AM.png
Screen Shot 2018-09-30 at 1.22.06 AM.png


Homework Equations


Fd = 0.5 * Cd * rho * A * v^2

The Attempt at a Solution


I can't even guess what those graphs consist of...
 

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Please be precise with the terms that you are using. Here are some things to think about. You say
Dowon Kim said:
The first graph is a velocity graph of a basketball with respect to time, ...
I assume you mean the "vertical component of the velocity with respect to time ..."
Dowon Kim said:
The second graph is the deceleration from the air resistance, ...
What exactly do you mean by this? I think you really mean "acceleration". How do you interpret the plateau in the graph that shows the acceleration to be roughly constant between about 0.5 s and 1.0 s?
Dowon Kim said:
However, I don't know what kind of equation is behind that graph.
You quoted an equation under "relevant equations". Can you solve it? After solving it, is there a way you can you determine whether it is relevant or not? In other words can it explain your data?
 
kuruman said:
Please be precise with the terms that you are using. Here are some things to think about. You say

"I assume you mean the "vertical component of the velocity with respect to time ...""
-Yeah, I do mean that. My bad.
"What exactly do you mean by this? I think you really mean "acceleration". How do you interpret the plateau in the graph that shows the acceleration to be roughly constant between about 0.5 s and 1.0 s?"
-Yes, it is acceleration, and I know that that constant region is due to the decreasing drag force when the basketball reaches the top of its projectile and therefore its velocity decreases.
"You quoted an equation under "relevant equations". Can you solve it? After solving it, is there a way you can you determine whether it is relevant or not? In other words can it explain your data?"
-It's the equation that I used to plot those graphs. Fd is the drag force, and all the variables other than v^2 are 0.0243, and v varies with time as two forces are acting on the basketball: gravity and air resistance.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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