A golf ball rolls off the top of a flight of concrete steps of total vertical height 4.00 m. The ball hits four times on the way down, each time striking the horizontal part of a different step 1.00 m lower. If all collisions are perfectly elastic, what is the bounce height on the fourth bounce when the ball reaches the bottom of the stairs?

The vertical height of the total no. of steps is\(h = 4\;{\rm{m}}\).

The width of each step is\(d = 1\;{\rm{m}}\).

In a perfectly elastic collision, the linear momentum of the system and its total energy remains conserved.

Here, the golf ball bounces back from every step with no loss in energy. The coefficient of restitution is given as:

\(e = \frac{{{\rm{Relative}}\;{\rm{velocity}}\;{\rm{of}}\;{\rm{separation}}}}{{{\rm{Relative}}\;{\rm{velocity}}\;{\rm{of}}\;{\rm{approach}}}}\)

For a perfectly elastic collision,\(e = 1\)such that

\(\begin{aligned}{c}1 = \frac{{{v_{\rm{f}}}}}{{{v_{\rm{i}}}}}\\{v_{\rm{f}}} = {v_{\rm{i}}}\end{aligned}\)

This implies that the golf ball bounces at the same speed. Therefore, it rises to the same height after every bounce, i.e., 4 m.