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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.


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.



Grade:

1 Answers

AKASH GOYAL AskiitiansExpert-IITD
419 Points
10 years ago

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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