Projectile motion of tennis ball with air resistance

In summary: If you select the coordinate axis upwards, the velocity is negative when the ball goes up. The contribution of drag is positive in this case. In summary, the conversation discusses the modeling of a tennis ball's path when serving, taking into account air resistance using differential equations. The equations for acceleration and drag coefficient are provided, with the correct sign for the y-equation being discussed. The lack of a closed form solution is mentioned and suggestions for solving the problem are provided. The importance of correctly choosing the sign for the drag term is also highlighted.
  • #1
Shruf
2
0
Hi, so I am trying to model the path of a tennis ball when serving. I already have the model without air resistance, but now I'm getting into differential equations with the air resistance. I obtained two differential equations for acceleration that i think are correct, but I'm not sure where to go from here exactly. It is in two dimensions. What I want to find is what angle the ball must be hit at to travel a set distance.


So far these are what i have:
equation 1: m[itex]\stackrel{d^{2}x}{dt^{2}}[/itex] = -k [itex]\stackrel{dx}{dt}[/itex] [itex]\sqrt{\stackrel{dx}{dt}^{2}+\stackrel{dy}{dt}^{2}}[/itex]

equation 2: m[itex]\stackrel{d^{2}y}{dt^{2}}[/itex] = k [itex]\stackrel{dy}{dt}[/itex] [itex]\sqrt{\stackrel{dx}{dt}^{2}+\stackrel{dy}{dt}^{2}}[/itex] - g

K is drag co-efficient, m is mass, g is gravity acceleration, all of which are defined.I have the initial velocity, and I know that I can use trig ratios to get the initial x and y velocities, but I have not idea what to do with them. BTW all the above below things are meant to be fractions, but i am not sure how to make them fractions, sorry.
 
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  • #2
Shruf said:
Hi, so I am trying to model the path of a tennis ball when serving. I already have the model without air resistance, but now I'm getting into differential equations with the air resistance. I obtained two differential equations for acceleration that i think are correct, but I'm not sure where to go from here exactly. It is in two dimensions. What I want to find is what angle the ball must be hit at to travel a set distance.


So far these are what i have:
equation 1: m[itex]\stackrel{d^{2}x}{dt^{2}}[/itex] = -k [itex]\stackrel{dx}{dt}[/itex] [itex]\sqrt{\stackrel{dx}{dt}^{2}+\stackrel{dy}{dt}^{2}}[/itex]

equation 2: m[itex]\stackrel{d^{2}y}{dt^{2}}[/itex] = k [itex]\stackrel{dy}{dt}[/itex] [itex]\sqrt{\stackrel{dx}{dt}^{2}+\stackrel{dy}{dt}^{2}}[/itex] - g

K is drag co-efficient, m is mass, g is gravity acceleration, all of which are defined.I have the initial velocity, and I know that I can use trig ratios to get the initial x and y velocities, but I have not idea what to do with them. BTW all the above below things are meant to be fractions, but i am not sure how to make them fractions, sorry.

Assuming drag is proportional to the square of the speed (which is a fair approximation at high speeds) then your equations are basically right. You do have the sign wrong on the drag term in the y-equation. Drag will always oppose the direction of movement.
For the LaTex, put the m inside the LaTex, use \frac, not \stackrel, and wrap the dx/dt terms in braces so that the squaring applies to the whole term, not just the x or y:
[itex]m\frac{d^{2}x}{dt^{2}}[/itex] = -k [itex]\frac{dx}{dt}[/itex] [itex]\sqrt{{\frac{dx}{dt}}^{2}+{\frac{dy}{dt}}^{2}}[/itex]
As to solutions, I believe there is no closed form solution.
 
  • #4
haruspex said:
As to solutions, I believe there is no closed form solution.

Thank; I do know that there is no closed solution but I am wondering how to do the mass calculations on something like excel? I am not sure how to incorporate these two equations for acceleration in my position equation, or in my calculations for range. Also, I think my sign is correct since the ball is being hit downwards, which I've made negative. I think i have to integrate the equations, but I'm not sure in what way I need to use them yet.
 
Last edited:
  • #6
Shruf said:
I think my sign is correct since the ball is being hit downwards, which I've made negative.
The contribution of drag to the acceleration must be a negative multiple of the speed.
 

Related to Projectile motion of tennis ball with air resistance

What is projectile motion?

Projectile motion is the motion of an object through the air that is affected only by gravity and air resistance. It follows a curved path called a parabola.

How does air resistance affect the trajectory of a tennis ball?

Air resistance, also known as drag, is a force that opposes the motion of an object through the air. For a tennis ball, air resistance slows down the ball and changes its trajectory, making it fall shorter and faster than it would without air resistance.

What factors affect the air resistance of a tennis ball?

The air resistance of a tennis ball is affected by its velocity, surface area, and shape. A faster moving ball with a larger surface area and a less streamlined shape will experience more air resistance.

How can the air resistance of a tennis ball be calculated?

The air resistance of a tennis ball can be calculated using the drag equation, which takes into account the density of the air, the velocity of the ball, its cross-sectional area, and a drag coefficient specific to the object's shape.

How does air resistance impact the distance a tennis ball travels?

Air resistance decreases the distance a tennis ball can travel compared to the same ball in a vacuum. The higher the air resistance, the shorter the ball's flight and the more it will drop during its flight.

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