A baseball is dropped from the top of a two storey building with the velocity of v2. At the same time a stone is thrown upward at a velocity of v1. If the magnitudes of the velocities of the two objects are equal, will the stone and ball meet up half way, above half way or below half way? Explain

The displacement covered by the baseball is modelled by the equation:

s_b = v_2t + \frac12 at^2

This is simplified to:

s_b = v_2t - 5t^2

Let's take g = 10m/s^2

The one for the stone is given by

s_s = v_1t -5t^2

Now, v1 = -v2

This gives:

s_s = -v_2t -5t^2

The two meet when they cover the same displacement, that is when sb = ss.

-v_2t -5t^2= v_2t - 5t^2

5t^2 cancels out and we get v_2t = 0

Which means, they only meet once, what is before they were thrown and they never meet again until they both stop falling. This is because they experience the same acceleration.