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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 ?
Taking gravity = 10m/s^2Lets 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 = v1applying formula v^2=u^2 +2 a Sv1^2 = 0 + 2 x 10 x 10v1 = root of (200) = 14.14Let velocity of ball just after leaving the ground = v2applying formula again v^2=u^2 +2 a S0 = v2^2 – 2 x 10 x 2.5v2 = root (50) = 7.07Notice that v2 and v1 are in different directionsNow applying formula v = u + atv2 = -v1 + a x .0114.14 = -7.07 + .01 x a1414 = -707 + aacceleration = 1414 + 707 = 2121 m/s^2
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