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vipul agrawal Grade:
        


A person jumps onto a cement floor from a height of 1 m and comes to rest in 0.1 second .the same person on jumping from a height of 9 m into a sand pit, comes to rest in 1 second . Compare the forces exerted on him by cement floor and sand pit.



6 years ago

Answers : (1)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

Dear Vipul


in first case


velocity v1 =√2gh1


chnage in momentum Δp1=mv1


force F1=Δp1/Δt1


in second case


velocity v2 =√2gh2


chnage in momentum Δp2=mv2


force F2=Δp2/Δt2


F1/F2=(Δt2*√h1)/(Δt1*√h2)=(1*1/0.1*3)=10/3


 


All the best.


AKASH GOYAL


AskiitiansExpert-IITD


 


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6 years ago
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