a 0.2kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds up at 25 m/s. If the ball is in contact with the sidewalk for 0.025s, what is the magnitude of the average force applied by the sidewalk on the ball?

Question

a 0.2kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds up at 25 m/s. If the ball is in contact with the sidewalk for 0.025s, what is the magnitude of the average force applied by the sidewalk on the ball?
I need steps on how to do this including equations ASAP!

Henry · Answer

a=V-Vo)/t = (-25-30)/0.025=2200m/s^2. F = m*a = 0.2 * 2200 = 440 N. Note: The velocity when leaving sidewalk is in opposite direction(neg.).