Answer Both objects will hit and collision certain regardless of velocity (providing it is greater than zero)
Ignore air resistance
Assume uniform gravitational field - ie constant acceleration
Reason
There are two ways of proving this, one involves maths. I will show you the easier way.
To see this you have to visualise the situation through the 'perspective' of object A
Object A is released at the same time Object B is fired. Both objects will accelerate downwards at g. Now the question is what would object A 'see'.
Since both objects accelerate at the same rate. With respect to object A, object B will not be accelerating. Eg imagine Two cars drag racing if their accelerating is the same there will be stationary with respect to each other.
now in this situation, object B has a initial velocity. Using the other analogy this would be equivalent to two cars drag racing both with the same acceleration(constant) but one started with initial velocity, thus it one driver will see the other car moving away at a constant velocity.
Apply this idea to this problem. This is what you will see if you were object A
You would see object B travelling at a constant velocity, because it is initially aimed at you, it will be guaranteed to it you. ( you are over a bottomless pit)
Side note
What you are doing is viewing the situation through a different frame of reference
in this frame of reference object A is stationary and object B is moving in a straight line at a constant velocity and the earth's surface is accelerating upwards at g.
Here is where i am a bit unsure
You should also note that none of these frames of reference available are travelling very fast and the gravitational field is relatively weak.
also note that events viewed from our frame of reference
( earth surface stationary) and the frame of reference of object A are equivalent.
Don't you agree that object A hits object B, at the same moment in time and location in space, when viewed from Earth's frame of and the falling frame of reference (object A's).
Just say we make this a rule ( it is quite intuitive)
Now imagine the situation if we replace object B with a mass-less particle (photon of light) which is projected a c. Do you see something strange?
In the falling frame of reference the same thing happens as discussed above the photon travels in a straightline.
From the earth's frame of reference the photon still travels in a straight line! and since object A will drop!. So between the frames of reference the time of collision agrees but but not the location!.
This could be a conundrum i would have to do the maths....( i hate maths) but i see on path. Don't you think this indicates a relationship between gravity, space and time ( for the rule to hold) ?
