Question

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))`.

Answer 1

Let `m` be the mass `rho` be the density of the ball. Therefore, density of water `=rho`.
Energy of the ball on striking the water surface `=mgh=mgxx19.6J` ..(i)
Net force opposing the motion of the ball in water
`=` upthrust `-` weight of ball `=((m)/(rho)xx2rhoxxg)-mg=mg`
If `d` is depth upto which the ball goes in water, then Work done `=` energy of the ball on striking water surface
`mgxxd=mgxx19.6,:.d=19.6m`
Velocity on striking the water surface `u=sqrt(2gh)=sqrt(2xx9.8xx19.6)=19.6m//s`
From `upsilon=u +at,` `0=19.6-(9.8)t:.t=2s`
This is the time taken by ball to travel a depth `(d)` in water.
Total time taken by the ball to reach the water surface `=2+2=4s`