A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. it faces an average retardation equal to (g= acceleration due to gravity): 1)g2)9g 3)100g 4)1000g

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Arun Kumar · Answer

Hi v^2=2gs=>200 again v^2=2as=>a=200/2=100m/s^2 Thanks & Regards Arun Kumar Btech, IIT Delhi Askiitians Faculty

KAPIL MANDAL · Answer

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A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. it faces an average retardation equal to (g= acceleration due to gravity): 1)g2)9g 3)100g 4)1000g

A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. it faces an average retardation equal to (g= acceleration due to gravity):
1)g2)9g
3)100g
4)1000g

2 Answers

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A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. it faces an average retardation equal to (g= acceleration due to gravity): 1)g2)9g 3)100g 4)1000g

A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. it faces an average retardation equal to (g= acceleration due to gravity):1)g2)9g3)100g4)1000g

2 Answers

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A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. it faces an average retardation equal to (g= acceleration due to gravity):
1)g2)9g
3)100g
4)1000g