# Projectile motion of tennis ball with air resistance

1. Dec 28, 2013

### Shruf

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.

2. Dec 28, 2013

### haruspex

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.

3. Dec 29, 2013

### ehild

4. Dec 29, 2013

### Shruf

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: Dec 29, 2013
5. Dec 29, 2013

### ehild

Last edited: Dec 29, 2013
6. Dec 30, 2013

### haruspex

The contribution of drag to the acceleration must be a negative multiple of the speed.