Describing motion from each frame of reference:
Since observer A is on one of the train's cars, he will find that he is at rest with respect to the train even when it is pulling away because, he is also moving with respect to the train.
Observer B on the stationary platform will observe that the train is pulling away from the station towards the right of the platform. as is described
Observer C will notice that the train is approaching him in the opposite direction at a speed which is the sum of the speeds at which both the train are traveling
If the conductor applies brakes on the train, since the platform is a stationary frame of reference. The motion will be observed as a simple decelerationg. he will observe that the force due to the brakes will cause the velocity of the train with respect to the train to decrease
Universal law of gravitation effects all objects alike . There will be a constant force of gravity acting on the train that will keep the train on its tracks. The tracks in turn exert a reaction force on the train. There will not be any affect on the train's motion as such( assuming that the train is moving along the tracks and gravitational force is exactly perpendicular)
Answer:
The answer to your question is: v = 121.46 m/s
Explanation:
Data
h = 30 m
vo = ?
dx = 300 m
Formula
t = 
speed = distance / time
Process
t= 2.47 s
speed = 300 / 2.47
speed = v = 121.46 m/s
Then, it jumps to HIGHER ORBITALS
