# Graph a missile with an acceleration normal and proportional to its velocity under effect of gravity

1. Oct 16, 2014

### powerof

1. The problem statement, all variables and given/known data

Title might be unclear since I had little space to explain. I'll state my problem from the beginning: I don't know how to determine one of the constants in formula for the velocity (the maximum velocity on the y axis to be precise). I need that constant because I want to make a plot without choosing a random value.

The problem statement (in my words because it's originally in a different language):

We'll develop a simple model for the trajectory of a tennis ball hit with a certain effect. The player hits the ball at an angle θ and in such a way that the ball spins creating an acceleration perpendicular and proportional to its velocity. We ignore friction with the air. Get the trajectory of the ball and compare it to the spin-friction-free model for ball (the parabolic one).

$v_0=\sqrt(v_{0_{x}}^{2}+v_{0_{y}}^{2})=30~m/s \leftarrow$ initial speed
$y_{0}=1~m\leftarrow$ height at which the ball is hit
$x_{0}=0\leftarrow$ we set the origin under the ball at the moment of impact
$\theta=30º=\pi/9 ~ rad\leftarrow$ angle at which the ball is hit
$(\mathbf{a}_{ef} \propto \mathbf{v}) ~\wedge~ (\mathbf{a}_{ef}\perp \mathbf{v} )\leftarrow$acceleration produced by the spin effect is perpendicular and proportional
$K=0.2$

2. Relevant equations

The equation of a projectile only under the effect of gravity with respect to which I have to compare my final solution:

$y(x,\theta)=y_{0}+(\tan \theta)\cdot x-\frac{g \cdot{(\sec \theta)}^{2}}{2\cdot{v_{0}}^{2}}\cdot x^{2}$

We can also get it as a function of time by setting $x(t)=v_{0_{x}} \cdot t$

3. The attempt at a solution

We can have two accelerations: inwards or outwards. This depends on the direction the ball is spinning. I'll only solve the problem for one case only because I can get the other one by switching the sign of the constant K.

$\textbf{a}_{T} = \textbf{a}_{e}+\textbf{a}_{g}=(-K \cdot v_{y},K \cdot v_{x})+(0,-g)=(-K \cdot v_{y},K \cdot v_{x}-g)=(\frac{d}{dt}v_{x},\frac{d}{dt}v_{y})\Rightarrow \\ \Rightarrow \left\{ \begin{array}{l} \frac{d}{dt}v_{y}=K \cdot v_{x}-g \Rightarrow v_{x}=\frac{1}{K}( \frac{d}{dt}v_{y}+g ) \\ \tfrac{d}{dt}v_{x}=-K \cdot v_{y} \Rightarrow \frac{d}{dt}(\frac{1}{K}(\frac{d}{dt}v_{y}+g) )=-K \cdot v_{y} \Rightarrow (\frac{d}{dt})^{2}v_{y}+K^{2}\cdot v_{y}=0 \end{array} \right.$

The solution of this equation is sinusoidal, of the form:
$$v_{y}(t)=C_{1}\cdot\sin(C_{2}\cdot t + C_{3})$$
Where $C_{i}$ are:

$C_{1}=v_{y_{MAX}}\leftarrow$ This is the amplitude, that is, the value which represents the maximum value the velocity reaches in the Y axis. I don't know what numerical value this has nor how can I get it from my initial data.
$C_{2}=(K^{2})^{1/2}=K\leftarrow$We get this by plugging the solution into the original differential equation
$C_{3}=\arcsin (\frac{v_{y_{0}}}{v_{y_{max}}})\leftarrow$ by setting t=0 and assuming C1 is correct, the equality follows

By plugging v_y into the first equation we get v_x and if we integrate both components of the velocity we get the trajectory (position function). Since I don't need the trajectory function right now and writing in LateX is a bit tedious for me, I'll stop here.

The tricky part was solving the differential equation and setting the constants right. Integrating the resulting velocity is easy and making the graph is easy as well provided I have correct values for all the constants.

Here comes my problem: I have no clue what v_y max is so I can't make a plot. I don't even know how to reason an approximation to it's value so that I can make an estimated guess to use when plotting.

Is there any way to deduce it without it being given as data?

English is not my first language so be patient if you find some weird wording or phrases.

Thank you for reading and have a nice day.

EDIT: Note that the problem statement doesn't ask me to plot the trajectory. This is only for my own curiosity. I want to plot it to see how it looks like with an additional perpendicular acceleration compared to gravitation only.

Last edited: Oct 16, 2014
2. Oct 21, 2014

### Greg Bernhardt

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Oct 21, 2014

### Staff: Mentor

Hi Powerof,
From the differential equations, for the velocity components, I get:

$$v_y=Asin(Kt)+Bcos(Kt)$$
$$v_x=\frac{g}{K}+Acos(Kt)-Bsin(Kt)$$

I obtain the constants of integration from the initial conditions on the velocity components.

Chet

4. Oct 21, 2014

### powerof

I think a post reword would be in order. Even I think it's a bit tedious to read so I'll try to cut off all the unnecessary parts. Is it possible to change the original post?

NEW TITLE: Help me determine one of the constants after solving a differential equation

1. The problem statement, all variables and given/known data

Develop a simple model for the trajectory of a tennis ball hit with a certain effect. The player hits the ball at an angle θ and in such a way that the ball spins creating an acceleration perpendicular and proportional to its velocity. We ignore friction with the air. Get the trajectory of the ball and compare it to the spin-friction-free model for ball (the parabolic one).

2. Relevant equations

The differential equation:

$\left\{ \begin{array}{l} \frac{d}{dt}v_{y}=K \cdot v_{x}-g \Rightarrow v_{x}=\frac{1}{K}( \frac{d}{dt}v_{y}+g ) \\ \tfrac{d}{dt}v_{x}=-K \cdot v_{y} \Rightarrow \frac{d}{dt}(\frac{1}{K}(\frac{d}{dt}v_{y}+g) )=-K \cdot v_{y} \Rightarrow (\frac{d}{dt})^{2}v_{y}+K^{2}\cdot v_{y}=0 \end{array} \right.$

Parabolic model of the throw:

$y(x)=y(x_{0})+\overset{y'(x_{0})}{\overbrace{\left(\tan\theta\right)}}(x-x_{0})+\overset{y''(x_{0})}{\overbrace{\left(\dfrac{a\sec^{2}\theta}{v_{0}^{2}}\right)}}\dfrac{1}{2}\left(x-x_{0}\right)^{2}$

3. The attempt at a solution

The solution of this equation is sinusoidal, of the form:
$$v_{y}(t)=C_{1}\cdot\sin(C_{2}\cdot t + C_{3})$$
Where $C_{i}$ are:

$C_{1}=v_{y_{MAX}}\leftarrow$ This is the amplitude, that is, the value which represents the maximum value the velocity reaches in the Y axis. I don't know what numerical value this has nor how can I get it from my initial data.
$C_{2}=(K^{2})^{1/2}=K\leftarrow$We get this by plugging the solution into the original differential equation
$C_{3}=\arcsin (\frac{v_{y_{0}}}{v_{y_{max}}})\leftarrow$ by setting t=0 and assuming C1 is correct, the equality follows

My problem is determining $v_{y_{MAX}}$ given that I know K, the initial velocity and angle and the initial positions.

Last edited: Oct 22, 2014
5. Oct 22, 2014

### powerof

That is most definitely easier than what I was doing. For some reason writing the solutions as $v_{y}=C_{1}\cdot\sin(C_{2}t+C_{3})$ and $v_{x}=\frac{g}{K}+C_{1}\cdot\cos(C_{2}t+C_{3})$ makes determining the constants harder.

In any case, thank you for your help. Knowing what the constants are will prove really useful when graphing the functions.