A ball of mass m is dropped onto a floor from a certain height. The...

Let the ball be dropped from a height $h$.

The velocity of the ball just before hitting the floor is $v = \sqrt{2gh}$.

Since the collision is perfectly elastic, the ball rebounds with the same speed $v$.

The change in momentum of the ball during one collision is $\Delta p = mv - (-mv) = 2mv$.

The time interval between two successive collisions is the time taken by the ball to go up to height $h$ and fall back to the floor, which is $T = \dfrac{2v}{g}$.

The average force exerted by the floor on the ball is $F_{avg} = \dfrac{\Delta p}{T} = \dfrac{2mv}{\dfrac{2v}{g}} = mg$.

By Newton's third law, the average force exerted by the ball on the floor is equal in magnitude, which is $mg$.