A ball is dropped on the floor from a height of 20 m . It rebounds to a height of 5 m . Ball remains in contact with floor for 1 s . The average acceleration during contact is (acceleration due to gravity =10 ~m / s ^2 )

Step-by-step solution

A ball dropped from h₁ = 20 m with g = 10 m/s ^2 reaches the ground with velocity v_ before = - 2gh₁ = - 2 10 20 = -20 m/s . After rebounding to h₂ = 5 m , its upward velocity is v_ after = 2gh₂ = 2 10 5 = 10 m/s . During the t = 1 s contact, the average acceleration is a_ avg = v t = v_ after - v_ before t = 10 - (-20) 1 = 30 m/s ^2 . The average acceleration is A .