Modeling a ball thrown vertically including drag from air resistance

In summary, the conversation discusses the modeling of a vertically thrown ball with a given starting velocity. The speaker has come up with a differential equation and is seeking confirmation on its correctness and alternative ways to solve it. Another speaker suggests simplifying the equation by using parameters and shows a simplified version. They also mention the possibility of using a u sub and an inverse trig table for solving the equation.
  • #1
MigMRF
15
0
So I'm trying to figure out how to model a ball getting thrown vertically with the starting velocity v_0. So I've come up with a differential equation which I'm pretty sure is correct:
1576515377922.png

Where D is a constant. So far so good. My problem is solving this. This is my attempt:
1576515472543.png

1576515451893.png

And when i do this and isolate v(t) i get this:
1576515564441.png

First of all, can anyone confirm this? It should only be used between t=0 and when the graph cross the x-axis. Secondly: Is there another waay to solve this.
BTW. if anyone were curios the fuction looks like this (for m=0.05 D=0.00085 g=9.82 v_0=16)
1576515835423.png
 
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  • #2
MigMRF said:
So I'm trying to figure out how to model a ball getting thrown vertically with the starting velocity v_0. So I've come up with a differential equation which I'm pretty sure is correct:
View attachment 254204
Where D is a constant. So far so good. My problem is solving this. This is my attempt:
View attachment 254206
View attachment 254205
And when i do this and isolate v(t) i get this:
View attachment 254207
First of all, can anyone confirm this?

That looks right (although it has ##x## instead of ##t## in there), but it can be significantly simplified by cancelling terms. Try using the parameter ##\alpha = \sqrt{\frac{D}{mg}}##.

PS You could also then try ##\beta = \tan(g\alpha t)##. That will sort out the inverse tangents.

PPS can you show that your solution tends to ##v = v_0 - gt## as ##D \rightarrow 0##?
 
Last edited:
  • #3
PeroK said:
That looks right (although it has ##x## instead of ##t## in there), but it can be significantly simplified by cancelling terms. Try using the parameter ##\alpha = \sqrt{\frac{D}{mg}}##.

PS You could also then try ##\beta = \tan(g\alpha t)##. That will sort out the inverse tangents.

PPS can you show that your solution tends to ##v = v_0 - gt## as ##D \rightarrow 0##?
Well not from the math, but it pretty sure it does looking at it from a graphical point of view.
1576522748945.png

In this example I've plottet the same things as before but with an v_0-g*t graph and they are pretty close. When you increase the mass they get closer and closer (drag goes down) while an increase in drag makes them go apart (with drag getting v=0 before nondrag ofc)
 
  • #4
I simplified things down to:
$$v = \frac{v_0 - \frac g \alpha \tan(\alpha t)}{1 + \frac{v_0 \alpha}{ g} \tan(\alpha t)}$$
Where ##\alpha = \sqrt{\frac{Dg}{m}}##

Which might be a bit easier to work with.
 
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  • #5
PeroK said:
I simplified things down to:
$$v = \frac{v_0 - \frac g \alpha \tan(\alpha t)}{1 + \frac{v_0 \alpha}{ g} \tan(\alpha t)}$$
Where ##\alpha = \sqrt{\frac{Dg}{m}}##

Which might be a bit easier to work with.
Thanks! It looks better this way, that's for sure :)
 
  • #6
MigMRF said:
Thanks! It looks better this way, that's for sure :)

It's not hard to get from your function to that. It might be worth it to check I haven't made any mistakes.
 
  • #7
I will put it back up that you could also do a u sub where ##u=Dv## and ##a=\sqrt{mg}##. Then use an inverse trig table.
 

1. How does air resistance affect the trajectory of a ball thrown vertically?

Air resistance, also known as drag, acts in the opposite direction of motion and slows down the ball's vertical velocity. This causes the ball to reach a lower maximum height and take longer to reach the ground compared to a ball thrown without air resistance.

2. How can the drag force on a ball be calculated?

The drag force on a ball can be calculated using the equation FD = 1/2 * ρ * v2 * A * CD, where ρ is the density of air, v is the velocity of the ball, A is the cross-sectional area of the ball, and CD is the drag coefficient.

3. What factors affect the drag force on a ball?

The drag force on a ball is affected by the density of air, the velocity of the ball, the cross-sectional area of the ball, and the drag coefficient. These factors can change depending on the environment and the characteristics of the ball.

4. How does the drag coefficient of a ball affect its trajectory?

A higher drag coefficient means that the ball experiences more drag force, which causes it to slow down more quickly and reach a lower maximum height. A lower drag coefficient results in less drag force and a higher maximum height.

5. Can air resistance be ignored when modeling a ball thrown vertically?

No, air resistance cannot be ignored when modeling a ball thrown vertically. It is an important factor that affects the trajectory and must be taken into account for accurate results.

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