A ball is thrown upward from the ground with an initial speed of 26.7 m/s; at the same instant, a ball is dropped from a building 13.1 m high. After how long will the balls be at the same height?

Answer 1

Solve this equation for t:
(Thrown) Ball 1 height = (Dropped) Ball 2 height

26.7 t - 4.9 t^2 = 13.1 - 4.9 t^2

The acceleration terms, -4.9 t^2, cancel out

t = 13.1/26.7 = ___ seconds

Interesting result, and one that hints at Einstein's principle of equivalence.

In a free-falling coordinate system the dropped ball doesn't move, and the thrown ball reaches it at constant speed. Gravity seems to disappear

Answer 2

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