2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 90 MPH is a really slow muzzle velocity! $\endgroup$ – Olin Lathrop Jan 20 '14 at 14:55
6
$\begingroup$

This has actually been tested and presented on television: the projectile will indeed fall straight to the ground to the stationary observer.

However, you the shooter will view the projectile traveling with a speed of 40 m/s while the projectile sees you traveling with a speed of 40 m/s, thus the airspeed remains.

$\endgroup$
2
$\begingroup$

Assuming the bullet instantaneously accelerates to 40m/s, then from the reference point of someone standing on the ground it will appear to drop straight down. In most cases, this is a good approximation because most objects we deal with everyday do not move 40m/s (thus we can neglect the distance the bullet travels during it's acceleration from the explosion).

However, in this case, the bullet is fired from a train that is moving 40m/s. Thus while the bullet is being accelerated it will also be moving some distance backward (i.e. in the direction the train is moving) before stopping completely.

We can think about this even more clearly by considering what it looks like from the point of view of the person firing the gun. To him, it would seem like he is standing still and the world around him is moving away at 40m/s. Now he fires his gun. He will see the bullet traveling away from him at 40m/s as anyone firing a gun would expect. But, an observe on the ground would see the bullet drop straight down, and the guy firing moving away from the bullet at 40m/s.

$\endgroup$
1
$\begingroup$

In the real world, the turbulence behind the train would make this rather complex, but it is probably easier to imagine throwing an anvil out of the back of the train (the speed would be a better match than your hypothetical, very slow gun) - get the speed just right and it would indeed drop straight down, from the perspective of a stationary observer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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