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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 Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
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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