Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I built a potato gun and wanted to calculate the muzzle velocity. I remember from physics that I could run the numbers by calculating time from launch until landing. After pointing straight into the air and launching the potato was air-born for 8.2sec.

Without air resistance that comes out to a muzzle velocity of around 90mph which is probably consistent as we can't see the projectile leave the muzzle.

I would like to get a more accurate calculation of what that would be with air resistance. Assuming a spherical potato that is 4ounces with a 2" diameter. Launched at 90 degrees with a 8.2sec air time.

share|cite|improve this question
If you assume quadratic drag the vertical trajectory can be solved analytically. See for the gory details. – John Rennie Nov 28 '12 at 16:56
@JohnRennie You made me put "\right)" five times next to each other, thank you very much :| – Claudius Nov 28 '12 at 20:32
up vote 3 down vote accepted

The air resistance of a sphere is given by²

$$ F_{\textrm{drag}} = \frac{1}{2} \rho C_d A v^2 $$

$C_d$ is usually set to $0.1$ for a sphere¹, $A$ is the relevant surface area, that is, $\left(\frac{\textrm{diametre}}{2}\right)^2 \pi$, $\rho \approx 1.2\textrm{ kgm}^{-3}$³.

We can set up a 1-dimensional coordinate system with "up" being in the positive $s$ direction. Then we have

$$ F_{\textrm{grav}} = - mg $$


$$ F_{\textrm{drag}} = - \left(\frac{1}{2} \rho C_d A \equiv K \right) \frac{v^3}{\sqrt{v^2}} $$

as it acts in the direction opposite to $v$. I have introduced a constant $K$ to simplify notation. Then:

$$ -m g - K \frac{v^3}{\sqrt{v^2}} = m \frac{\textrm{d}v}{\textrm{d}t} $$

which is a first-order differential equation in $v$ but second-order in $s$. However, it is not trivial to solve it, as it contains terms non-linear in $v$. Solving the system numerically is also non-trivial, as one of the initial conditions, $v(0)$ is unknown (i.e. we can’t simply evolve it in time from $t = 0$).

The boundary/initial conditions we have are:

$$ s(t = 0) = 0 \quad ; \quad s(t=8.2) = 0 ; \quad v(t=0) = v_0 $$

I would probably go about setting $v(0)$ to some number, then let the system evolve and check where the object is at $t = 8.2$ - if $s$ is positive, decrease $v(0)$, if $s$ is negative, increase $v(0)$ (basically solve the problem numerically).

I think solving the problem numerically would be easier if we had the maximum height $s_{\textrm{max}}$ in the objects path. Setting the time at which it reaches this height to $0$, we would have:

$$ s(0) = s_{\textrm{max}} \quad ; \quad v(0) = 0 $$

and one could simply evolve the system backwards in time until $s(t) = 0$.

I am sorry I cannot give a complete answer, but maybe someone else has an idea on how to continue from here?


John Rennie posted a helpful link which claims to have an analytic solution to this problem. I did not verify said solution, but picked out two formulae:

$$ t_{\textrm{imp}} = \tau \cosh^{-1}\left( \exp\left( \frac{y_{\textrm{peak}}}{\tau v_t}\right) \right)$$

$$ y_{\textrm{peak}} = - v_t \tau \ln\left( \cos\left( \tan^{-1}\left( \frac{v_0}{v_t} \right) \right) \right) $$

where $\tau$ is the characterstic time, $v_t$ is the terminal velocity (the maximum velocity a freely-falling object reaches due to air drag opposing gravity) and $t_{\textrm{imp}}$ is the time after which an object reaches the ground again. $v_0$ is the inital velocity we’re looking for.

Rearranging this gives:

$$ \tan \left( \cos^{-1} \left( \exp \left( - \ln \left( \cosh \left( \frac{t_{\textrm{imp}}}{\tau} \right) \right) \right) \right) \right) v_t = v_0 $$

$v_t$ is given as

$$v_t = \sqrt{\frac{2 m g}{C_d \rho A}}$$

and $\tau = v_t / g$. Plugging all this together gives me:

$$ v_0 = 91.032\textrm{ ms}^{-1} = 203\textrm{ mph} $$

For reference, the thing I put into Qalculate is:

tan(acos(e^(−ln(cosh(8.2s × 9.81 N/kg / sqrt( (2× 4ounce × 9.81 N/kg)/(0.1 × 1.29 kg/m^3 × Pi × (1 in)^2)))))))×sqrt( (2× 4ounce × 9.81 N/kg)/(0.1 × 1.29 kg/m^3 × Pi × (1 in)^2)) 
share|cite|improve this answer
"maybe someone else has an idea on how to continue from here?" Computationaly. The traditional answer is that at this point you turn to a computer. For speeds on order of 100 MPH (i.e. about 160 KPH) and a tolerance for several percent model dependent uncertainty almost any quadrature will do. – dmckee Nov 28 '12 at 16:37
Solving this numerically appears to be the only way forward for me, too, but with the original data, one would have to solve an equation with two integrals numerically, which I find rather…ugly (although certainly do-able within a short time). I was hoping there was any way to reduce this sensibly to a first-order ODE or to massage our data somehow to "just" integrate twice rather than having to solve an equation with these integrals, too (as hinted at in the last part). – Claudius Nov 28 '12 at 16:41

As John Rennie points out, you can solve Claudius equation analytically is you split it into two different cases. Your basic equation is

$$m \dot{v} = -mg \mp k v^2,$$

with the drag being negative on the way up, and positive coming down. It will make things simpler to make everything dimensionless. We have $[m]=M$, $[k]=ML^{-1}$ and $[g]=LT^{-2}$, so we can construct the following dimensionless variables for time, space, velocity and acceleration:

$$t=\sqrt{\frac{m}{gk}}\tau,\ x=\frac{m}{k}\xi,\ v=\sqrt{\frac{gm}{k}}\omega,\ a=g\alpha,$$

and turn the equation into

$$\dot{\omega} = -1 \mp \omega^2,$$

or equivalently

$$\frac{\dot{\omega}}{1 \pm \omega^2} = -1.$$

On the way up, positive sign in the denominator, this can be integrated as

$$\arctan \omega = -\tau + T_1,\ \omega = \tan(T_1-\tau),$$

If the launch happens at $\tau=0$ with $\omega=\omega_0$, we can figure out that $T_1=\arctan \omega_0$, which is also the time it takes the projectile to climb to its apex..

Integrating one more time,

$$\xi = \log(\cos(\arctan \omega_0-\tau)) + \Xi_1,$$

and at $\tau=0$ we have $\xi=0$, so $\Xi_1, = -\log(\cos(\arctan \omega_0))$, or

$$\xi = \log \frac{\cos(\arctan \omega_0-\tau)}{\cos(\arctan \omega_0)},$$

and the maximum height reached by the projectile will be

$$\xi_{max} = \log \frac{1}{\cos(\arctan \omega_0)}.$$

On the way down, negative sign in the denominator, we can also integrate to get

$$\tanh^{-1}\omega = -\tau +T_2,\ \omega = \tanh(T_2-\tau)$$

and we want to have $\omega=0$ when $\tau=\arctan \omega_0$, to have a common times origin, so we again get $T_2 = \arctan \omega_0$.

Integrating once more,

$$\xi = -\log(\cosh(\arctan \omega_0-\tau)) + \Xi_2,$$

and since when $\tau=\arctan \omega_0$ we have $\xi=\log \frac{1}{\cos(\arctan \omega_0)}$, we can figure out $\Xi_2 = \log \frac{1}{\cos(\arctan \omega_0)}$, so

$$\xi = -\log(\cosh(\arctan \omega_0-\tau)) + \log \frac{1}{\cos(\arctan \omega_0)},$$

and when your potato returns to the ground $\xi=0$ and

$$\cosh(\arctan \omega_0-\tau) = \frac{1}{\cos(\arctan \omega_0)}.$$

From this last equation, you want to figure out $\omega_0$, your launch speed, given $\tau$, the total flight time. And this last part, I'd suggest doing numerically...

share|cite|improve this answer
I find it curious that this equation doesn’t reduce to the one I used for $\{v,\omega\}_0$, at least not trivially. – Claudius Nov 28 '12 at 22:18
Oh, there is a more than reasonable chance that I messed up somewhere in between... $y_peak$ is equivalent, so it should have happenned in the derivation of the trip down, will try to figure it out later... – Jaime Nov 28 '12 at 23:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.