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ICSE Class 10 Answered

A body of mass 20 kg falls from a height of 78.4 m to the ground and rebounds to a height of 19.6 M if the ball made contact with the ground for one fifth of a second hand force in KGF exerted by the ground on the body
Asked by Samrat | 30 Jun, 2018, 09:12: PM
answered-by-expert Expert Answer
velocity v1 of the body when it reaches ground, v1 =begin mathsize 12px style square root of 2 cross times g cross times h end root equals space square root of 2 cross times 9.8 cross times 78.4 end root space equals space 39.2 space m divided by s end style
Rebound velocity V2 for the body to reach a height 19.6 m,  v2 = begin mathsize 12px style square root of 2 cross times g cross times h apostrophe end root space equals space square root of 2 cross times 9.8 cross times 19.6 end root space equals space 19.6 space m divided by s end style
force = begin mathsize 12px style fraction numerator C h a n g e space i n space M o m e n t u m over denominator t i m e end fraction space equals space fraction numerator m left parenthesis v 1 plus v 2 right parenthesis over denominator t end fraction space equals space fraction numerator 20 cross times left parenthesis 39.2 plus 19.6 right parenthesis over denominator begin display style bevelled 1 fifth end style end fraction space equals space 5880 space N end style 
( velocities are in opposite direction. hence change in momentum is calculated as m(v1+v2) )
Answered by Thiyagarajan K | 02 Jul, 2018, 12:29: PM

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