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about you mass 20 kg falls from a height of 78.4 to the ground and rebounds to the height of 19.6 on the ball and mAde contact with the ground for one fifth of the second find the force in kilogram force exerted by the ground on the body The answer in my book is 2000 kgf
Asked by abhyudaysinghbhadauriya | 24 Jan, 2020, 01:40: AM
answered-by-expert Expert Answer
velocity of the object when it is hitting the ground  v1 = (2gh)1/2 = (2 × 9.8 × 78.4 )1/2 = 39.2 m/s
 
 
rebound velocity of the object after hitting the ground, v2 = (2gh)1/2 = ( 2 × 9.8 × 19.6 )1/2 = 19.6 m/s
 
Change in momentum = m( v1 - v2 ) = 20 × ( 39.2 + 19.6 ) = 58.8 ( kg m/s )
 
If contact time with ground is 0.2 seconds,
then Force exerted by ground = Change in momentum / time = 58.8 / .2 = 294 N
Answered by Thiyagarajan K | 24 Jan, 2020, 09:08: AM

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