Question

a tennis ball is dropped on the floor from height 20 m. it rebounds to a height of 5 m. the ball was in contact with the floor for 0.101 s what was its average acceleration during the contact​

Answer 1

Explanation:

when it is dropped from 10m Initial height 10m initial velocity 0 velocity just before hotting ground v2gh V2 9.8 10 14.07 m/s (downward) after rebound, maximum height reached 2.5m final velocity at top 0 initial velocity(just after rebound) V2gh v2*9.8*2.5 V49 7 m/s (upward) assuming downward as positive direction So velocity just before hitting ground +14.07 m/s velocity just after hitting ground -7 m/s change in velocity +14.07 - (-7) 21.07 m/s time 0.01s acceleration change in velocity/time 21.07/0.01 2107 m/s2 2100m/s2

Answer 2

this is the process put your values and get your ans.