# What was the muzzle velocity of a home made gun launch strait in the air if air time was 8.2sec?

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 strait 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.

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If you assume quadratic drag the vertical trajectory can be solved analytically. See hyperphysics.phy-astr.gsu.edu/hbase/mechanics/quadrag.html 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

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

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

and

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

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?

Update

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))

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"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...

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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