Projectile motion of tennis ball with air resistance

AI Thread Summary
The discussion focuses on modeling the trajectory of a tennis ball during a serve, incorporating air resistance through differential equations. Two equations for acceleration are presented, but there is uncertainty about their correctness, particularly regarding the sign of the drag term in the vertical equation. It is noted that drag should oppose the direction of movement, and the equations need to be properly formatted for clarity. The user seeks guidance on integrating these equations to calculate the angle required for the ball to travel a specific distance. There is an acknowledgment that no closed-form solution exists for this problem, and suggestions for using Excel for calculations are mentioned.
Shruf
Messages
2
Reaction score
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\stackrel{d^{2}x}{dt^{2}} = -k \stackrel{dx}{dt} \sqrt{\stackrel{dx}{dt}^{2}+\stackrel{dy}{dt}^{2}}

equation 2: m\stackrel{d^{2}y}{dt^{2}} = k \stackrel{dy}{dt} \sqrt{\stackrel{dx}{dt}^{2}+\stackrel{dy}{dt}^{2}} - 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.
 
Physics news on Phys.org
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\stackrel{d^{2}x}{dt^{2}} = -k \stackrel{dx}{dt} \sqrt{\stackrel{dx}{dt}^{2}+\stackrel{dy}{dt}^{2}}

equation 2: m\stackrel{d^{2}y}{dt^{2}} = k \stackrel{dy}{dt} \sqrt{\stackrel{dx}{dt}^{2}+\stackrel{dy}{dt}^{2}} - 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:
m\frac{d^{2}x}{dt^{2}} = -k \frac{dx}{dt} \sqrt{{\frac{dx}{dt}}^{2}+{\frac{dy}{dt}}^{2}}
As to solutions, I believe there is no closed form solution.
 
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:
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.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top