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

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,

`$dP = F * dT$`

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 * [V_f - V_o] $`

= 82 x (1.1 - 5.5)

= -360 kg

∴ the net force acting will be

`$F=(dP)/(dT)$`

`$F=(-360)/(1.65)$`

= 218 N