newtonian mechanics - Bouncing ball time problem

I'm studying a problem and I encountered a strange problem:

When a ball bounces how much time does the ball spend while touching the floor?

To be more clear I suppose that when a ball bounces the actual bounce can't start EXACTLY when the ball touches the floor but the ball touches the floor, the Energy from the fall is given to the floor then the floor gives the energy back and the ball bounce; but in what time?

Answer

An possible (simplistic) answer would be the following: a simple model for the bouncing ball is a spring that shrinks to absorb all the initial kinetic energy and restores fully. To put it into equations, call $v_0$ the initial velocity of the ball, $m$ its mass and $K$ the spring stiffness. The initial kinetic energy is $\frac12mv_0^2$. If the spring shrinks by a length $x$, the elastic energy is $\frac12Kx^2$. The mechanic energy is conserved, so we have all along the move $$\frac12mv^2+\frac12Kx^2=\frac12mv_0^2.$$ Let us express the velocity $v$ as a function of $x$ during the first half of the move (when the spring shrinks and the ball slows down) $$v=\sqrt{v_0^2-\frac Kmx^2}.$$ Now remark that $v=\frac{\mathrm dx}{\mathrm dt}$, so we can separate the $x$ variable and get the differential $$\frac{\mathrm dx}{\sqrt{v_0^2-(K/m)x^2}}=\mathrm dt$$ and we can integrate this equation from the moment to ball hits the ground to the moment it stops (the $\frac12$ coefficient is there because it is only one half of the movement) $$ \frac12T=\int_0^{x_0}\frac{\mathrm dx}{\sqrt{v_0^2-(K/m)x^2}}=\frac\pi2\sqrt{\frac mK}$$ (with $x_0=v_0\sqrt{m/K}$). So the result is $$\boxed{T=\pi\sqrt{\frac mK}.}$$ Interestingly, it does not depend on the initial velocity !

A more refined model would take into account the spherical shape of the ball. If it has a Young modulus $E$, the elastic energy could be (very roughly) approximated by $\frac{2\pi}3 Ex^3$ (this means that the more the ball is shrinked the more it resists to an extra shrinking). The result is $$ T=2.19187\left(\frac m{E v_0}\right)^{1/3}.$$ Now it depends on the initial velocity. (The constant is equal to $6^{1/3}\pi^{1/6}\Gamma(4/3)/\Gamma(5/6)$).

Of course the total deformation of the ball should be taken into account, the preceding approach is valid only for hard balls (large $E$). Note also that dissipation of energy has not been taken into account.