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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.
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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