A ball is droped on to the floor from a height of 10 m it rebounds to a height of 2.5 m if a ball is in contact with floor for 0.01sec what is the average acceleration
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The answer is g/3=9.8/3=3.26 m/s²
Because during free fall the accelaration is 'g' but when it hits the floor it applies a force on it and in response the floor applies an equal and opposite force on the ball but this force acts for only 0.01 s and hence the ball rebounds. But the moment the ball leaves the floor the only force acting is the gravity only so the acceleration is again 'g'.
But during the contact with the floor the ball is at rest for 0.1s. It means net force acting on it is 0N. During this time the accelaration is 0.
Hence, Average accelaration=(g+0+g)/3=g/3
Because during free fall the accelaration is 'g' but when it hits the floor it applies a force on it and in response the floor applies an equal and opposite force on the ball but this force acts for only 0.01 s and hence the ball rebounds. But the moment the ball leaves the floor the only force acting is the gravity only so the acceleration is again 'g'.
But during the contact with the floor the ball is at rest for 0.1s. It means net force acting on it is 0N. During this time the accelaration is 0.
Hence, Average accelaration=(g+0+g)/3=g/3
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