Ask a New Question

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!
12 years ago

Answers

Henry
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.).
12 years ago

Related Questions

a rubber ball is dropped from a height of 5m. After the 5th bounce, the ball only comes back up to 0... A rubber ball dropped from a height of exactly 5 ft bounces (hits the floor) several times, losing 1... A rubber ball dropped from a height of exactly 6 ft bounces several times on the floor, loosing 10%... A rubber ball is dropped from the top of a hole.Exactly 3.0 seconds later, the sound of the rubber b... A rubber ball is dropped onto a ramp that is tilted at 20 degrees. A bounding ball obeys the "law o... A rubber ball is dropped from the top of a ladder. It bounces on the same spot on the ground several... 17. A rubber ball dropped on a hard surface takes a sequence of bounces, each one 3 / 5 as high as... A rubber ball is dropped from a height of 9m and each time it strikes the ground, pounces to a heigh...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use