Computing velocity as a function of time

Question

I have measured the amount of time (5 ms) it takes a projectile to travel 1 meter. I make the following assumptions:

• $1$-D motion.
• the only force acting on the projectile is a force due to drag, $F(t)$.
• the force due to drag is simply proportional to the square of the velocity, e.g., $F(t) = Cv(t)^2$.

Do I have enough information to compute the velocity as a function of time? Thanks!

Answer 1

Let's think about the information you have. From Newton's 2nd Law, you can say that $m\dot v\hat v=-cv^2\hat v$. This is an easy ODE to solve, but you don't have any information about the initial velocity, so this method won't work. However, you can also write Newton's 2nd Law in terms of position, that is, $m\ddot x\hat x=-c\dot x^2 \hat x$. This is a second order ODE, and you have two initial conditions for the position, $x(0\ \mathrm{ms})=0\ \mathrm{m}$, and $x(5\ \mathrm{ms})=1\ \mathrm{m}$. Can you take it from here?

Answer 2

You will need to know the mass $m$ of the projectile (you can weigh it), and with your current data you will need to know what $C$ is. You can either do this by

• Looking up the drag coefficient for the projectile, measuring the cross sectional area of the projectile and the density of the medium it is going through.

• Making another simultaneous measurement (i.e. measuring the amount of time it takes to make another distance).

In either case, we wish to solve the ODE $$x''(t)=-\frac{C}{m}(x'(t))^2,$$ where $x$ is the position. Substituting $v=x'$ and solving the ODE by separation of variables gives $$v(t)=\frac{m}{Ct-K_1 m},$$ where $K_1$ is a constant to be determined. Substituting back and integrating gives $$x(t)=\frac{m}{C}\log|Ct-K_1 m|+K_2,$$ where $K_2$ is another constant to be determined. We have the boundary data $x(0)=0$ and $x(0.005)=1$. From the first boundary condition, we obtain $K_2=-\frac{m}{C}\log|K_1 m|$. This results in $$x(t)=\frac{m}{C}\log\left|\frac{Ct-K_1 m}{K_1 m}\right|.$$ The second boundary condition results in the condition $$\left|\frac{0.005 C-K_1 m}{K_1 m}\right|=e^{C/m},$$ which has two solutions $$K_1\in \left\{-\frac{0.005C}{m(e^{C/m}-1)},\quad \frac{0.005C}{m(1+e^{C/m})}\right\},$$ but the first solution is the only physical solution (otherwise the initial velocity is negative, and the velocity will diverge at a certain time). Hence the solution is given by $$x(t)=\frac{m}{C}\log\left(200(e^{C/m}-1)t+1\right), \tag{1}$$ $$v(t)=\frac{200m(e^{C/m}-1)}{200C(e^{C/m}-1)t+C}.$$ You now need to know $C$. The first approach is straightforward (i.e. use the well-known equation $F=\frac{1}{2}\rho C_D A v^2$). If you are opting for the latter approach, you can measure the time it takes to make another distance (e.g. $2$ metres) to obtain a third boundary condition and then solve for $C$ in $(1)$. For this, you can use a numerical root-finding method such as the Newton-Raphson method.