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

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