Have a pneumatic cannon that launches a mouse via a pusher plate when lured into the barrel. How many Gs will the mouse feel when launched?
Pusher Plate Stroke = 8"
Distance Mouse Travels (at 45 deg): 20 feet
Weight of Mouse: 0.8oz (is this needed?)
This is a real thing that has attracted some criticism for being cruel to animals, even though it is one of many ways to collect vermin for disposal. I'd like to be able to point out that the G forces are non-lethal. The mouse/rat is in contact with the pusher plate at the time of firing so it isn't smacked in the face with the pusher...it's pushed. I figured that if a human cannonball can be fired for 200 feet uninjured, what about this mouse being shot 20 feet into a bucket?
Let's first solve for the launch velocity. If the landing is at the same level as the cannon, the range of a projectile (assuming no air drag etc) is $$ d = \frac{v^2}{g} \sin{2 \theta} $$ where $v$ is the launch velocity, $g$ is the acceleration of gravity and $\theta$ is the launch angle. Solving for $v$, we obtain $$ v_{\text{launch}} = \sqrt{ \frac{dg}{\sin{2\theta}} } $$ Now let's look a the the barrel. It's essentially a linear path that the projectile goes through before launching into the air. The initial velocity is zero, so the following equation ca be written for the velocity of the projectile within the barrel: $$ v(t) = at \qquad \Rightarrow \qquad t = \frac{v_{\text{launch}}}{a} $$ where $a$ is a (constant) acceleration and $t$ is time. On the other hand, integrating this equation, we can obtain the equation for the distance (as a function of time) inside the barrel: $$ s(t) = \frac{1}{2}at^2 $$ plugging in the previous result for time, we obtain $$ s(t) = \frac{1}{2} a \left( \frac{v_{\text{launch}}}{a}\right)^2 = \frac{ v^2_{\text{launch}} }{2a} $$ from which we can solve for the acceleration $a$: $$ a = \frac{ v^2_{\text{launch}} }{2s} $$ Here, $s$ is the length of the barrel. Now we can plug in the result for the launch velocity: $$ a = \frac{ \left( \frac{dg}{\sin{2\theta}} \right) }{2s} = \frac{dg}{2s \, \sin{2\theta}} $$ the "G" acceleration is specifically acceleration divided by $g$, so the "G" acceleration as a formula is $$ \frac{a}{g} = \frac{d}{2s\, \sin{2 \theta}} = \frac{ \left(6.096~\text{m} \right)}{2 \left(0.203~\text{m} \right) \sin{90^{\circ}}} \approx 15~\text{g's} $$