Question

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

Answer 1

Considering the jump to the pool,

$v^2 = u^2 + 2gh \newline where \space u=0 \newline h= height \space of \space the \space tower \newline v= final \space velocity$

$g =9.81m s^{-2} \newline \because v=\sqrt{2gh} \newline v=\sqrt{2*9.81*18} \newline =\underline{\underline{18.79ms^{-1} }} \\~\\ This \space v \space acts \space as \space u' \space for \space deaccelaration \space in \space the \space water \\~\\ For \space the \space water, \newline v'^2 = u'^2 -2as \newline where\space v'=0 , \newline u'=18.79ms^{-1} \newline s=2.2m \\~\\ solving \space this \space gives \\~\\ a=\underline{\underline{80.24ms^{-2}}} \\~\\ Substituding \space to \space F=ma \newline where \space m= 80kg \newline F=80*80.24 \newline =\underline{\underline{6419.2N}}$