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Grade: 11

                        

A ball dropped on to the floor from a height of 10m rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.02s , its average acceleration during contact is

2 years ago

Answers : (1)

Khimraj
3007 Points
							
initial velocity = \sqrt{2gh} = \sqrt{2*10*10} = 10\sqrt{2}
final velocity = \sqrt{2gh} = \sqrt{2*10*2.5} = 5\sqrt{2}
change in velocity = 10\sqrt{2} -(- 5\sqrt{2}) = 15\sqrt{2}
average acceleration = 15\sqrt{2}/0.02 = 750\sqrt{2} m/s2
Hope it clears........................................
2 years ago
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