Question

Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before striking the water her speed is 5.50M/S at a time of 1.65 after entering the water her speed is reduced to 1.10M/S. What is the force that acts on her when she is in the water

Answer 1

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o$ = 5.50 m/s

Her speed after entering the water, $V_f$= 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times dT$

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

    $= m \times [V_f - V_o]$

    = 82 x (1.1 - 5.5)

    = -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

  = 218 N