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

`19.6m, 4 sec`

Explanation

Velocity of ball at the instant striking the water `v=sqrt(2gh)`
`v=19.6m//s`
Let `v` is the volume and `P` is the density of ball
`F=B-W`
`F=V2rhog-V.rho.g`
`F=Vrhog`
`F=mg`
`a=guarr`
`(i) d = (v^(2))/(2a) " "implies d = (19.6xx19.6)/(2xx9.8) implies d = 19.6m`
`(ii) T = (2u)/(a) " "implies T = (2xx19.6)/(9.8) = 4sec`.