Average force on a bouncing ball

A massive clue comes from being asked for force and given time . There are only two places in physics where those two interact: (1) power, which would require work, and the force here is not applied over a distance or (2) impulse, which only requires the change in momentum to be found. Newton's second law in terms of impulse is

$F_{avg}=\frac{\Delta p}{\Delta t}.$

In order to find $\Delta p$ , use conservation of energy. First, for the way down: the ball is dropped from a height of 5 $\text{m}$ to the ground below, so

$U_i=K_f.$

Cleverly choose the kinetic energy definition in terms of momentum to produce the equation

$mgh=\frac{p^2}{2m}$

$p=\sqrt{2m^2gh}$

$p=\sqrt{2(1)^2(10)(5)}=\sqrt{100}=10\text{ kg m/s}.$

Similarly, for the rebound after the ball bounces,

$K_i=U_f$

$\frac{p^2}{2m}=mgh$

$p=\sqrt{2m^2gh}$

$p=\sqrt{2(1)^2(10)(1.8)}=\sqrt{36}=6\text{ kg m/s}.$

Finally, double back to Newton's second law above:

$F_{avg}=\frac{\Delta p}{\Delta t}$

$F_{avg}=\frac{p_f-p_i}{\Delta t}.$

Since the one momentum is down and the other is up, one needs to be negative in order to indicate their antiparallel relationship.

$F_{avg}=\frac{10-(-6)}{0.1}=160\text{ N}.$