Conceptual Physics (12th Edition)

As far as vertical motion is concerned, the steel ball started from rest, and drops 0.8 m (from the tabletop into the circle defining the top of the can). Find the time needed to drop that distance. $$0.8 m = \frac{1}{2}g t^{2} = 5 \frac{m}{s^{2}}t^{2}$$ $$t = 0.4 s$$ This is discussed on page 48. The horizontal velocity of the steel ball does not change. Find the horizontal distance it moved in the 0.4 s drop time. $$d = vt = (8.0 \frac{m}{s})(0.4 s) = 3.2 m$$