The period is twice the time it takes to fall from height $h$ starting from rest. $$\frac{1}{2}gt_{\rm fall}^2 = h$$ $$t_{\rm fall} = \sqrt{2h/g}$$ $$ T = 2\sqrt{2h/g}$$ The speed at which the ball hits the ground can be got by conservation of mechanical energy: $$\frac{1}{2} mv_b^2 = mgh$$ $$ v_b = \sqrt{2gh}$$ $$\Delta p = 2m\sqrt{2gh}$$ $$F_{\rm avg} = \frac{\Delta p}{T} = \frac{ 2m\sqrt{2gh}}{ 2\sqrt{2h/g}}$$ $$ F_{\rm avg} =\frac{ \cancel{2}m\sqrt{\cancel{2}g\cancel{h}}}{ \cancel{2}\sqrt{\cancel{2}\cancel{h}/g}} = mg$$ The average force by the ball on the floor is equal in magnitude and it is the same as when the ball is resting on the floor.
The average force is equal to the impulse divided by the time interval between bounces. Measure the time interval between the two highest points of the bounce. The starting height is h and by conservation of energy the next highest point it reaches is ηh. \begin{aligned} \Delta t =&~ t_{\rm fall} + t_{\rm rise}\\ =& \sqrt{2h/g} + \sqrt{2\eta h/g}\\ =& \sqrt{2h/g}(1+\sqrt{\eta}) \end{aligned} The change in momentum is equal to the impulse: \begin{aligned} \Delta p =&~ p_{\rm f} - p_{\rm i}\\ =&~ m v_{\rm f}-mv_{\rm i}\\ =& -m \eta v_{\rm i}-mv_{\rm i}\\ =& \sqrt{2g\eta h} +\sqrt{2g h}\\ =&\sqrt{2g h}(1+\sqrt{\eta}) \end{aligned} Finally divide impulse by time interval to get the average force. \begin{aligned} F_{\rm avg} =& \frac{\Delta p}{\Delta t}\\ =& \frac{\sqrt{2g h}(1+\sqrt{\eta})}{\sqrt{2 h/g}(1+\sqrt{\eta})}\\ =& mg \end{aligned} A not completely unexpected result.