Question #100518

A Man of mass 80 kg dives into a swimming pool from a tower of height 18 m. He was found to go down in water by 2.2 m and then started rising. Find the average resistance of water. Neglect the air resistance of water take value of, g =9.81m/s2

Expert's answer

Considering the jump to the pool,

"v^2 = u^2 + 2gh \\newline\nwhere \\space u=0 \\newline\nh= height \\space of \\space the \\space tower \\newline\nv= final \\space velocity"

"g =9.81m s^{-2} \\newline\n\\because v=\\sqrt{2gh} \\newline\nv=\\sqrt{2*9.81*18} \\newline\n=\\underline{\\underline{18.79ms^{-1} }} \\\\~\\\\\n\nThis \\space v \\space acts \\space as \\space u' \\space for \\space deaccelaration \\space in \\space the \\space water \\\\~\\\\\n\nFor \\space the \\space water, \\newline\nv'^2 = u'^2 -2as \\newline\nwhere\\space v'=0 , \\newline\nu'=18.79ms^{-1} \\newline\ns=2.2m \\\\~\\\\\nsolving \\space this \\space gives \\\\~\\\\\na=\\underline{\\underline{80.24ms^{-2}}} \\\\~\\\\ Substituding \\space to \\space F=ma \\newline\nwhere \\space m= 80kg \\newline\nF=80*80.24 \\newline\n=\\underline{\\underline{6419.2N}}"

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