Solution :
Given weight of Kathy = 82 kg
Her speed before striking the water, [tex]$V_o $[/tex] = 5.50 m/s
Her speed after entering the water, [tex]$V_f$[/tex]= 1.1 m/s
Time = 1.65 s
Using equation of impulse,
[tex]$dP = F \times dT$[/tex]
Here, F = the force ,
dT = time interval over which the force is applied for
= 1.65 s
dP = change in momentum
dP = m x dV
[tex]$= m \times [V_f - V_o] $[/tex]
= 82 x (1.1 - 5.5)
= -360 kg
∴ the net force acting will be
[tex]$F=\frac{dP}{dT}$[/tex]
[tex]$F=\frac{-360}{1.65}$[/tex]
= 218 N
