Average force for repeated elastic collisions

Here's how. The ball gains a velocity $v$ due to gravity before hitting the ground. So each time it hits the ground its velocity is changed from $v$ to $-v$ (taking down as positive) during the collision, then returning again with $v$.

The force $F_1= m\frac {dv}{dt}$ is experienced by the ball due to the collision, however this force is felt after the weight gets cancelled out by the normal contact force. So we have $$F_N=F_1+W$$ $$\Rightarrow F_N=m\frac{dv}{dt}+mg$$ If the total time taken was $t$, where $t$ is the sum of the duration of all collisions, then the sum of the forces experienced would be (assuming you know basic calculus) $$\int_0^t F_N\,dt=m\int_0^t \frac{dv}{dt}\,dt+\int_0^t mg\,dt$$ $$\Rightarrow \int_0^t F_N\,dt=m\int_v^{-v} dv+\int_0^t mg\,dt$$ $$\Rightarrow \int_0^t F_N\,dt=2mv+mgt$$ Hence the average force would be $$F_{avg}=\frac{\int_0^t F_N\,dt}{t}=\frac{2mv+mgt}{t}=\frac{2mv}{t}+mg $$ as $t \rightarrow \infty , \frac1t\rightarrow0,\;$ $$\therefore F_{avg}=mg$$ Notice, if this kind of bouncing happened horizontally (e.g. a ball colliding back and forth between two walls) then the $F_{avg}$ would be zero, just as you had thought. Hope this helps.