an object is dropped from a height of 10m. it rebounds to a height of 2.5m . if it is in contact with the floor for 0.01 sec, then what is the acceleration during contact ?

Question

Manas Shukla · Answer

Taking gravity = 10m/s^2 Lets first find out the change in velocity of the ball while its in contact with the ground . So , Let Velocity of the ball just before contact with the ground = v1 applying formula v^2=u^2 +2 a S v1^2 = 0 + 2 x 10 x 10 v1 = root of (200) = 14.14 Let velocity of ball just after leaving the ground = v2 applying formula again v^2=u^2 +2 a S 0 = v2^2 – 2 x 10 x 2.5 v2 = root (50) = 7.07 Notice that v2 and v1 are in different directions Now applying formula v = u + at v2 = -v1 + a x .01 14.14 = -7.07 + .01 x a 1414 = -707 + a acceleration = 1414 + 707 = 2121 m/s^2

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an object is dropped from a height of 10m. it rebounds to a height of 2.5m . if it is in contact with the floor for 0.01 sec, then what is the acceleration during contact ?

an object is dropped from a height of 10m. it rebounds to a height of 2.5m . if it is in contact with the floor for 0.01 sec, then what is the acceleration during contact ?

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