Question

A solid ball of density half that of water,falls freely under gravity from a height of 19.6 m and then enters water upto what depth will the ball go

Answer 1

Let density of ball is ρ
Then density of water is 2ρ [ as question said density of ball is half of density of water ]
Let mass of ball is m , then volume of ball is m/ρ

Now, on striking water surface , potential energy of ball = mgh = mg × 19.6m [ as height is 19.6m ]
Due to buoyancy ,
Net force opposes the motion of ball = upthrust - weight of ball
= (Volume of ball × density of water × g) - mg
= m/ρ × 2ρ × g - mg = 2mg - mg

let ball goes h depth in water.
then potential energy of ball in h depth = mg × h

Use wrok energy theorem,
potential energy of ball on water surface = potential energy of ball in h depth of water
mg × 19.6m = mg × h
h = 19.6

Hence, answer should be 19.6 m