Can a bullet leave a gun and tumble to the ground?

This question seems to have been asked a few times in different configurations, but none of them answer my variation. I've struggled to understand this for nearly 15 years and had conflicting answers from my school physics teachers and more recently friends who are physicists.

So a round leaves my gun barrel at $40$ m/s. Its initial Airspeed is $40$ m/s.

In another question they asked about firing down the length of a train traveling the same speed. I understand that firing toward the front of the train would result in a ground speed of $80$ m/s and an airspeed of $40$ m/s. Also, firing toward the back of the train results in a ground speed of $0$ m/s and an airspeed of $40$ m/s.

My question... if you were to walk to the back of the train, open the door and fire directly out the back, would you end up with a ground and airspeed of $0$ m/s? Meaning the projectile would literally just tumble in a straight trajectory down to the ground?

Plenty of people have thrown spanners into this one over the years - like talking about the way an explosive force will 'hang' in a certain space if not pushed, pushing the projectile away from itself as well as the firing pin of the gun, giving it extra forward momentum in that instance. (unlike firing from the front of the train would always equal $80$ m/s for air and ground. The explosion cannot be 'left behind'). I personally don't buy this one... But i don't know enough to judge.

• 90 MPH is a really slow muzzle velocity! – Olin Lathrop Jan 20 '14 at 14:55