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For comparison what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.0-ms-long encounter with a hard floor?
Wiki User
∙ 15y ago
Its velocity just before it hits the floor is found from kinematic equations and acceleration due to gravity; V = SqRt{2gy) = SqRt{19.6} = 4.4 m/s . The velocity after 1 ms is zero. The average deacceleration is the velocity difference divided by the time during impact; a = (4.4 - 0 )/.001 = 4400 m/ss .
Wiki User
∙ 15y ago
The acceleration experienced by the test tube would be approximately 980 m/s^2, which is equivalent to the acceleration due to gravity acting on the object when it falls freely. This acceleration would occur over the short 1.0-ms interval during the impact with the hard floor.
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Zero is.
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