To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00m. It rebounds to a height of 2.00m. If the ball is in contact with the floor for 12.0ms, a) what is the magnitude of its average acceleration during that contact and b) is the average acceleration up or down? i used the equation x-x0 = v0t + 1/2 at^2. to find the time when the tennis ball just touches the ground, i did 4 = 4.9t^2, assuming down in positive and up is negative. so t = .904. then i used the equation v = v0 + at to find the velocity when the ball just touches the ground. i did v = (9.8)(.904) = 8.85m/s. the ball was in contact with the floor for 12.0ms so the final time is .904 + .012 = .916. so i have my final and initial times and i get .012ms when i subtract them. the only thing i'm missing to find the average acceleration is the final velocity. i tried using v = v0 +at again, v = 8.85 + (9.8)(.916) = 17.826m/s. so i get that the average acceleration is 748.06 upwards. but the answer in my book says 1260. how did they get that answer?