Question

Heather drops a ball weighing 0.50kg a distance of 2.0m. If the speed of the ball right after it bounces is the same speed of the ball right before it bounces, and the interaction time between the ball and the floor was 0.10 seconds, what was the average force of the floor on the ball?

Answer 1

The speed of the ball just before impact was v=√(2gh) = 6.26m/s.  The acceleration is twice this over the time (twice because the second speed is the same in the other direction, meaning the total change in speed is 2V)
a = 12.52/0.10 = 125.2m/s²
The force is F=ma, so F = 0.5kg·125.2m/s² = 62.6N