A soild 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. How much time will it take to come again to the water surface? Neglect air resistandce and viscosity effects in water. (Take `g=9.8 m//S^(2))`.

Explanation

Suppose m is mass of the ball and `rho` is density of the ball. Therefore, density of water `=2rho.`
As the ball falls from a height of 19.6m, therefore, energy of the ball on striking the water surface `=mgh=mgxx19.6J`.
Net force opposing the motion of the ball in water.
= upthrust - wight of ball
`=(m)/(rho)xx2rhoxxg-mg=mgN`
If d is depth upto which the ball goes in water, then
work done `=` energy of the ball
`mgxxd=mgxx19.6:. d=19.6m`
Velocity on striking the water surface
`v=sqrt(2gh)=sqrt(2xx9.8xx19.6)=19.6m//s`
If t is time taken by the ball to travel a depth (d) in waer, then from
`v=u+at`
`0=19.6-(9.8)t, t=(19.6)/(9.8)=2s`
Total time taken by the ball to come again to the water surface `=2+2=4s.`