In this problem, the ball’s kinetic energy will be equivalent to the potential energy at the initial and final points of each rising or falling cycle.

Given data

The height from which the ball is dropped is\(h = 1.60\;{\rm{m}}\).

The rebound height is \(h' = 1.20\;{\rm{m}}\).

The fraction of potential energy is given by the following:

\(\frac{{{\rm{P}}{{\rm{E}}_{\rm{f}}}}}{{{\rm{P}}{{\rm{E}}_{\rm{i}}}}} = \frac{{mgh'}}{{mgh}}\)

Here, m is the mass of the ball, and g is the gravitational potential energy of the ball.

Plugging the values in the above equation,

\(\begin{array}{l}\frac{{{\rm{P}}{{\rm{E}}_{\rm{f}}}}}{{{\rm{P}}{{\rm{E}}_{\rm{i}}}}} = \left[ {\frac{{1.20\;{\rm{m}}}}{{1.60\;{\rm{m}}}}} \right]\\\frac{{{\rm{P}}{{\rm{E}}_{\rm{f}}}}}{{{\rm{P}}{{\rm{E}}_{\rm{i}}}}} = 0.75\end{array}\)

The number of rebounds the ball can be calculated as the following:

\(\begin{array}{c}{\left( {\frac{{{\rm{P}}{{\rm{E}}_{\rm{f}}}}}{{{\rm{P}}{{\rm{E}}_{\rm{i}}}}}} \right)^n} = \left( {90\% - 100\% } \right)\\{\left( {\frac{{{\rm{P}}{{\rm{E}}_{\rm{f}}}}}{{{\rm{P}}{{\rm{E}}_{\rm{i}}}}}} \right)^n} = 10\% \end{array}\)

Plugging the values in the above equation,

\(\begin{array}{c}{\left( {{\rm{0}}{\rm{.75}}} \right)^n} = \frac{{10}}{{100}}\\n = \frac{{\log 0.1}}{{\log 0.75}}\\n \approx 8\end{array}\)

Thus, \(n \approx 8\) is the required number of rebounds made by the ball.