A ball of relative density 0.8 falls into water from a height of 2 m. find the depth to which the ball will sink (neglect viscous forces)

Explanation

Method - 1
Initial velocity of the body when it touches the water surface `v_(1)` is given by `v_(1)^(2)-0^(2)=2gh rArr v_(1)=sqrt(2gh)`
Retardation of the ball moving in downard direction inside the water due to buoyancy force
`-a=("upthrust"-"weight")/("mass")=(V_("total")(rho_(w))g-V_("total")(rho_(b))g)/(V_("total")(0.8))`
`=((1)g-(0.8)g)/((0.8))rArr -a=(g)/(4)`
distance travelled by the by body inside the water before coming to rest is given by
`0^(2)-v_(1)^(2)=2ah^(1)" "(h^(1)-"distance travelled in water")`
`h^(1)=(-v_(1)^(2))/(2a)=(2gh)/(2g//4)=4h=8rArr h^(1)=8m`
Method - 2
Gravitational potential energy = Apparent weight of the body `xx` Displacement of the body inside liquid
`mgh=mg^(1)h^(1) ("where "g^(1)=g((1)/(d//rho_(w))-1)`)
`"Relative density"=(d)/(rho_(w))=0.8`
`g^(1)=g((1)/(0.8)-1)=g((0.2)/(0.8))=(g)/(4)`
`h^(1)=(g)/(g^(1))g=(g)/(g//4)xx2=8m`