A solid ball of density half that of water falls freely under gravity from a height of 19.6 m and then enters water. Neglecting air resistance and viscosity effect in water, the depth up to which the ball will go is `(g= 9.8 m//s^(2))`

Options

A. 19.6 m
B. 9.8 m
C. 14.7 m
D. 12.7 m

Answer

A

Explanation

Velocity of ball on reaching the water surface is `v=sqrt(2 gh)=sqrt(2xx9.8xx19.6)=19.6 m//s`
Let `rho` be the density of ball, then density of water is `2 rho`.
If a is the reatardation of the ball in water, then
`a=("upward thrust - weight")/(mass) =(V(2rho)g - V(rho)g)/(V rho) =g=9.8 ms^(-2)`
The distance through which the ball goes in water is `S=(v^(2))/(2a) =((19.6)^(2))/(2xx9.8)= 19.8 m`