Question - I am trying to work out the time taken for an American football launched from head height at an angle $\theta$ to the horizontal to hit the ground. I am involving air resistance. However, I am struggling to work out how to work out / assign a value to a constant $k$. Here is my working:
I let $\mathbf{r}$ denote the position vector of the ball, so $\mathbf{\dot r}$ denotes the velocity. $\mathbf{u}$ denotes the crosswind force / velocity of air. So air resistance must be $-k(\mathbf{\dot r} - \mathbf{u})$ as it will be proportional to the difference in velocity between the ball and air.
I then use $\mathbf{F}=m\mathbf{a}$ to give $m\mathbf{\ddot r} = m\mathbf{g}-k(\mathbf{\dot r} - \mathbf{u})$.
This gives the differential equation $\mathbf{\ddot r} + \frac{k}{m}\mathbf{\dot r} = \mathbf{g} + \frac{k}{m}\mathbf{u}$.
I solved this to give $\mathbf{r} = \mathbf{C_1} + \mathbf{C_2}e^{\frac{-kt}{m}}+(\frac{m}{k}\mathbf{g}+\mathbf{u})t$.
The initial conditions must be $\mathbf{r}(0) = \mathbf{H}$ as the ball is launched from a displacement $\mathbf{H}$ above the ground. Also, $\mathbf{\dot r}(0) = \mathbf{V}$ as the ball is launched with an initial velocity $\mathbf{V}$.
This gives $\mathbf{H} = \mathbf{C_1} + \mathbf{C_2}$ and $\mathbf{V}=-\frac{k}{m}\mathbf{C_2}e^{\frac{-kt}{m}} + \frac{m}{k}\mathbf{g} + \mathbf{u}$.
Solving I got $\mathbf{C_1}=\mathbf{H}+\frac{m}{k}\mathbf{V}-\frac{m^2}{k^2}\mathbf{g}-\frac{m}{k}\mathbf{u}$ and $\mathbf{C_2} =-\frac{m}{k}\mathbf{V}+\frac{m^2}{k^2}\mathbf{g}+\frac{m}{k}\mathbf{u}$.
So the particular solution is $\mathbf{r} = \mathbf{H}+\frac{m}{k}\mathbf{V}-\frac{m^2}{k^2}\mathbf{g}-\frac{m}{k}\mathbf{u} + (-\frac{m}{k}\mathbf{V}+\frac{m^2}{k^2}\mathbf{g}+\frac{m}{k}\mathbf{u})e^{\frac{-kt}{m}} + (\frac{m}{k}\mathbf{g} + \mathbf{u})t$.
After factorising I have: $\mathbf{r} = $ $\mathbf{H} + \frac{m}{k}(\mathbf{V}-\frac{m}{k}\mathbf{g}-\mathbf{u}+(-\mathbf{V}+\frac{m}{k}\mathbf{g}+\mathbf{u})e^{\frac{-kt}{m}})+(\frac{m}{k}\mathbf{g} + \mathbf{u})t$
So, $\mathbf{r}=$ $\mathbf{H}+ \frac{m}{k}(\mathbf{V}-\frac{m}{k}\mathbf{g}-\mathbf{u})(1-e^{\frac{-kt}{m}})+(\frac{m}{k}\mathbf{g} + \mathbf{u})t$.
Now notice that the vectors can be written as: $\mathbf{V} = V \cos{\theta}\mathbf{i}+ V \sin{\theta}\mathbf{k} $, $\mathbf{g} = -g{\mathbf{k}}$, $\mathbf{u} = u\mathbf{j}$.
This gives: $\mathbf{r} = H\mathbf{k} + \frac{m}{k}\left(V\cos{\theta}\mathbf{i} + V\sin{\theta}\mathbf{k}+\frac{mg}{k}\mathbf{k}-u\mathbf{j}\right)\left(1-e^{\frac{-k}{m}t}\right)+\left(-\frac{mg}{k}\mathbf{k}+u\mathbf{j}\right)t$.
Now I will solve for $t$ and work out when ball hits ground
The ball hits the ground when the $\mathbf{k}$ coefficient is $0$. So I have $H + \frac{m}{k}(1-e^{\frac{-k}{m}t})V \sin{\theta} + \frac{m}{k}(1-e^{\frac{-k}{m}t})\frac{mg}{k} + -\frac{mg}{k} = 0$.
My issue / problem
I have $H=1.79, g=9.81, m = 0.43, V = 14.3$. I choose $\theta = 55^\circ$ but I guess this could be any reasonable value. How do I find out $k$ though ?
I understand that there isn't a set value for this as it depends on the ball. Can anyone give me a reasonable value for $k$ for an American football ? I used a value of $0.07$ because I let $mg=kv$ where $v$ is the terminal velocity of the ball to solve for $k$. However, this gave me a value for time $t$ of $2.8$ seconds. I guess I need to know the terminal velocity of an American football / rugby ball - does anyone know this value ?
I know $2.8$ seconds is way too small, so my question is if there is anything I have done wrong in my working in setting the model up and solving ? I would really appreciate this help
The $\mathbf{k}$ coefficient is $$\frac{m \left(1-e^{-\frac{k t}{m}}\right) \left(\frac{g m}{k}+V \sin (\theta )\right)}{k}-\frac{g m t}{k}+H=0$$
The solution for the flight time $t_{flight}$ is:
$$t_{flight}\to \frac{m W\left(-\frac{(g m+k V \sin (\theta )) e^{-\frac{k (H k+m V \sin (\theta ))}{g m^2}-1}}{g m}\right)}{k}+\frac{H k}{g m}+\frac{V \sin (\theta )}{g}+\frac{m}{k}$$
with $W(z)$ the Lambert W-Function (ProductLog).
Substituting your numeric values we get $t_{flight} \simeq 2.41338$, but not $2.8$
The $\mathbf{i}$ coefficient is
$$\text{x}(t)\to \frac{m V \cos (\theta ) \left(1-e^{-\frac{k t}{m}}\right)}{k}$$
Hence the horizontal distance travelled is
$$x(t_{flight}) \simeq 16.3693$$