train and arrow

[marchez]

if i was on a train travelling at 100mph, and i stood at the back of the train and fired an arrow out of the window at 100mph, what would the people on the staion see??

[Peter Pedant]

a bullet going at 200 mph

This is galilean relativity.

Premsumably you were firing the arrow in the direction of travel and not at right angles.

If it is at right agles, then it follows a parabola and will hit you on the nose when it returns.

[marchez]

i am firing the arrow in the opposite direction of travel

[Ray The Grey]

It would hit you when it returns ? It is an arrow not a boomerang... It would not travel back on itself - the parabola would arc forwards and away from the train, if fired at right angles.

The people on the platform would see an arrow fly out of the rear window relatively slowly and land quite near behind the train. The disturbance of air and the momentum of air behind the train would mean the arrow travelled a short distance and then be drawn back and hit the ground. A lot of the air is moving at a similar speed to the train even if the spectators aren't. The air turbulence effect would be hard to estimate though.

[Bonzo 2000]

I dont agree with either, I'm afraid!  If the train is moving one way at 100mph and you fire the arrow at 100mph in the opposite direction, the net speed of the arrow would be zero.  The punters on the platform would see the arrow in its bow in your hands in the train travelling at 100mph, until the point when the arrow is released, at which point it would stop dead and drop straight to the ground.  Obviously the air turbulence at the back of a speeding train would stir things up a bit, but effectively the arrow would stop and drop.

[El D]

incorrect - the arrow possesses an aerodynamic shape combined with more energy due to its larger mass focused on a point greater than that of the air opposing it. it would go forward a little then drop as already suggested. speeds are secondary to energies involved.

[Hamish]

What are you going on about energy? Kinetic energy =0.5mv^2 by vector combination energy at final arrow speed defore deceleration (-100mph) therefore KE=0.5*m*(v1-v2)^2 but v1=v2 therefore KE=0.

<br><br>

Observers will see arrow moving in direction of train but sowing down as arrow accelerates (v1>v2) at 100mph (v1=v2) arrow appears stationary. Assuming equal decelaration after peak v2, ie v1 and v2 decrease at equal rates, arrow will drop. However once the arrow has left the bow it is no longer reciving any force in the direction of the train and therefore v1 will reach zero before v2 and the arrow will land a litte forward of the position it reaches when v1=v2.

<br><br>hope this helps

[drbsparks]

some idiot leaning out of the window