How long will you collide?

By conservation of energy

$\large \frac12mv^2+\frac12Kx^2=\frac12mv_0^2.$

Where $v_{0}=\sqrt{2gH}$

Thus

$\large v=\frac{dx}{dt}=\sqrt{v_0^2-\frac Kmx^2}.$

Solving for $t$

$\large \frac{\mathrm dx}{\sqrt{v_0^2-(K/m)x^2}}=\mathrm dt$

$\large \frac12T=\int_0^{x_0}\frac{\mathrm dx}{\sqrt{v_0^2-(K/m)x^2}}=\frac\pi2\sqrt{\frac mK}$

Where $x_{0}=v_0\sqrt{m/K}$

$\large \boxed{T=\pi\sqrt{\frac mK}.}$

If you consider the ball as a Ideal spring, then the effective mass will be 1/3 not 1.BEcause the mass 1kg becomes the mass of the spring, Not the mass of an object hanging on the end of the spring.

The system works like a oscillator, and the required time is half period. So:

t = T/2 = pi*sqrt(m/k) ~ 0.314

Let's say the ball just contacted the ground with $v_{0}$ at $t=0$ . From the eq. of motion $m \ddot{x} +kx=0$ , $x(t)=v_{0}\sqrt{\frac{m}{k}}\sin{(\sqrt{\frac{k}{m}}t}).$ Then, $\dot{x}(t)=0$ when $t=\frac{\pi}{2}\sqrt{\frac{m}{k}}$ . $T=\pi\sqrt{\frac{1}{100}}=\frac{\pi}{10}.$

kx=mgh; x=0.196m

v^2=u^2+2aS; v=(39.2)^1/2

t=x/t; hence, t=0.313

The f of spring = sqrt(k/m)/(2 * pi) .... Hence T = (2 * pi)/sqrt(k/m)
But the ball is only compressed, not stretched. That would take only half the period T... T/2 = pi/sqrt(k/m) = pi/sqrt(100/1) =. 314